Heat of reaction for the reaction is:

$PCl_5(g)+H_2O(g)\longrightarrow POCl_3(g)+2HCl(g)$

Given that,

$P_{white} + (3/2)Cl_2(g) +\frac{1}{2}O_2(g)\longrightarrow POCl_3; \: \Delta H=-135.5 \:kcal$

$H_2( g ) + Cl_2( g )\longrightarrow +2HCl(g); \: \Delta H=-44.1 \:kcal$

$P(w)+(5/2)Cl_2(g)\longrightarrow PCl_5(g); \: \Delta H=-89.6 \:kcal$

$H_2(g)+\frac{1}{2}O_2(g)\longrightarrow H_2O(g);\: \Delta H=-57.8\:kcal$

0%
$-32.2 kcal$
0%
$-52.5 kcal$
0%
$-45.2 kcal$
0%
None of these

Explanation

$P_{white} + \dfrac 32 Cl_2(g) +\dfrac{1}{2}O_2(g)\longrightarrow POCl_3; \: \Delta H_1=-135.5 \:kcal$

$H_2( g ) + Cl_2( g )\longrightarrow +2HCl(g); \: \Delta H_2=-44.1 \:kcal$

$P(w)+ \dfrac52 Cl_2(g)\longrightarrow PCl_5(g); \: \Delta H_3=-89.6 \:kcal$

$H_2(g)+\cfrac{1}{2}O_2(g)\longrightarrow H_2O(g);\: \Delta H_4=-57.8\:kcal$

The final reaction is obtained by reversing 3 and 4 and adding 1 and 2 to it:

$PCl_5(g)+H_2O(g)\longrightarrow POCl_3(g)+2HCl(g)$

$\Delta H = \Delta H_1 + \Delta H_2 - \Delta H_3 - \Delta H_4$ .

$\implies$

$\Delta H = - 135.5 - 44.1 + 89.6 + 57.8$

$=- 32.2 kcal$

The dissolution of 1 mole of NaOH(s) in

$100 \: mole \: of \: H_2O(l)$ give rise to evolution of heat as -42.34 kJ. However, if1 mole of NaOH(s) is dissolved in

$1000 \:mole \:of \:H_2O(l)$ the heat given out is 42.76 kJ.

If the enthalpy change when

$900 \:mole \:of \:H_2O(l)$ are added to a solution containing 1 mole of NaOH(s) in

$100 \:mole \:of \:H_2O$ is

$x$

$kJ$ :

0%
$-0.42$
0%
$0.42$
0%
$0.24$
0%
None of these

Explanation

$\Delta H_1 for \: 1 \: mole \: NaOH \: in \: 100 \: mole \: H_2O(l) = -42.34 \: kJ$

$NaOH(s) + 100H_2O\longrightarrow NaOH(100H_2O); \: \Delta H= -42.34 \: kJ$ ... (i)

$NaOH(s) + 1000H_2O\longrightarrow NaOH(1000H_2O); \: \Delta H= -42.76 \: kJ$ ... (ii)

By eqs. [(ii) - (i)]

$NaOH(s) + 900H_2O\longrightarrow NaOH(900H_2O); \: \Delta H= -0.42 \: kJ$

Hence,option A is correct.

The heat measured for a reaction in bomb calorimeter is :

0%
$\Delta G$
0%
$\Delta H$
0%
$\Delta U$
0%
$P\Delta V$

Explanation

A bomb calorimeter is a type of constant-volume calorimeter used in measuring the heat of combustion of a particular reaction.

Hence, the heat measured for a reaction in bomb calorimeter is

$\Delta U$ .

1 gram sample of

$NH_4NO_3$ is decomposed in a bomb calorimeter. The temperature of the calorimeter increases by 6.12 K. The heat capacity of the system is 1.23 kJ/g-deg. What is the molar heat of decomposition for

$NH_4NO_3$ ?

0%
-7.53 kJ/mol
0%
-398.1 kJ/mol
0%
-16.1 kJ/mol
0%
602 kJ/mol

Explanation

The expression for the heat evolved is

$q= mS\Delta T$

When 1g of ammonium nitrate decomposes, the heat evolved is

$q=1\times 1.23\times 6.12$

When 1 mole of ammonium nitrate decomposes, the heat evolved is

$q=1\times 1.23\times 6.12\times 80=6 02.21 \:kJ/mol$

The

$\Delta_f H^{\circ}$ for

$CO_2 (g), \: CO(g) \: and \: H_2O(g)$ are

$-393.5, -110.5$ and

$-241.8 \: kJ \: mol^{-1}$ respectively. The standard enthalpy change (in kJ) for the reaction is:

$CO_2 (g)+ H_2 (g)\longrightarrow CO(g)+ H_2O(g )$

0%
524.1
0%
41.2
0%
-262.5
0%
-41.2

Explanation

First write the balanced chemical reactions for the formation of carbon dioxide, carbon monoxide and water.

$C+ O_2 \longrightarrow CO_2; \: \Delta H^{\circ}= -393.5 \:kJ$ ... (i)

$C+\frac{1}{2}O_2\longrightarrow CO;\: \Delta H^{\circ}= -110.5 \:kJ$ ...(ii)

$H_2+\frac{1}{2}O_2\longrightarrow H_2O;\: \Delta H^{\circ}= -241.8 \:kJ$ ...(iii)

Then obtain the required equation by equating

$(ii) + (iii) - (i):$

$CO_2+H_2\longrightarrow CO+H_2O; \: \Delta H^{\circ}=+41.2$

Hence, the standard enthalpy change for the reaction is 41.2 kJ/mol.

Combustion of carbon studies in a bomb calorimeter to obtain heat of reaction. Which of the following options are correct?

0%
The value obtained shows change in heat enthalpy.
0%
The value obtained shows change in internal energy.
0%
The volume remains constant.
0%
The pressure remains constant.

The commercial production of water gas utilize the reaction under standard conditions:

$C+H_2O(g)\longrightarrow H_2(g)+CO$

The heat required for this endothermic reaction may be supplied by adding a limited amount of air and burning some carbon to

$CO_2$ . How many g of carbon must be burnt to

$CO_2$ to provide enough heat for the water-gas conversion of 100 g carbon?

Neglect all heat losses to the environment. Also

$\Delta _fH^{\circ}$ of CO,

$H_2O(g)$ and

$CO_2$ are -110.53, 241.81 and -393.51 kJ/mol respectively.

0%
$33.4 g$
0%
$66.72 g$
0%
$15.18 g$
0%
None of these

Explanation

$C+{ H }_{ 2 }O\left( g \right)\rightarrow { H }_{ 2 }+CO$

$\Delta { H }=\Delta { { H }_{ f } }\left( CO \right) -\Delta { { H }_{ f } }\left( { H }_{ 2 }O \right)$

$=\left\{ \left( -110.53 \right) -\left( -241.81 \right) \right\} kJ/mole=131.28kJ/mole$

so, total amount of energy to be provided for conversion of 100 gm of C

$\displaystyle(100gC)\left( \frac { 1mole C }{ 12.0gC } \right) \left( \frac { 131.28kJ }{ mol } \right) =1094kJ$

To provide that energy, amount of C to be burnt

$\displaystyle\left( 1094 \right) kJ\left( \frac { 1mole C }{ 393.51kJ } \right) \left( \frac { 12.0g C }{ mole C } \right) =33.4g$

The enthalpies of formation of

$N_2O$ and

$NO$ are

$28$ and

$90\ kJ mol^{-1}$ respectively. The enthalpy of the reaction,

$2N_2O(g) +O_2(g)\longrightarrow 4NO(g)$ is equal to :

0%
8 kJ
0%
88 kJ
0%
-16 kJ
0%
304 kJ

Explanation

First write the balanced chemical; equations for he formation of dinitrogen oxide and nitric oxide.

$N_ 2+\frac{1}{2}O_2\longrightarrow N_2O; \: \Delta H=28 \:kJ$ ...(i)

$\frac{1}{2}N_ 2+\frac{1}{2}O_2\longrightarrow NO; \: \Delta H=90 \:kJ$ ...(ii)

Then multiply second equation with 4 and first equation with 2.

$2N_ 2+O_2\longrightarrow 2N_2O; \: \Delta H=2 \times 28 \:kJ$ ...(iii)

$2N_ 2+2O_2\longrightarrow 4NO; \: \Delta H=4 \times 90 \:kJ$ ...(iv)

Now substract third equation from fourth equation.

$2N_2O +O_2\longrightarrow 4NO; \: \Delta H=304 \:kJ$ .

Thus,the enthalpy of the reaction

$2N_2O(g) +O_2(g)\longrightarrow 4NO(g)$ is 304 kJ.

Select the incorrect statements:

0%
A few combustion process are endothermic
0%
Heat of combustion may be positive
0%
Exothermic compounds arc more stable than endothermic compounds
0%
Hess's law can he verified experimentally

Which of the following are applicable for a thermochemical equations? It tells:

0%
about the physical state of reactants and products.
0%
about the allotropic form (if any) of the reactants.
0%
whether the reaction is exothermic or endothermic.
0%
whether a particular reaction is spontaneous or not.

Which of the following are correct for the given diagram?

0%
$\Delta H_3=\Delta H_1+\Delta H_2$
0%
$\Delta H_1=0$
0%
$\Delta H_2 = (C_{calorimeter}+ C_{product})\times (T_1-T_2)$
0%
$\Delta H_2 = (C_{calorimeter}+ C_{product})\times (T_2-T_1)$

Explanation

The following are correct for the given diagram.
(A)

$\Delta H_3=\Delta H_1+\Delta H_2$ . This is because enthalpy is a state function and its value depends on the initial and final states only. It is independent of the path followed.

(B)

$\Delta H_1=0$ as the process is adiabatic and no heat energy is supplied.

(C)

$\Delta H_2 = (C_{calorimeter}+ C_{product})\times (T_1-T_2)$ .
The enthalpy change is due the change in the temperature.

The following is incorrect for the given diagram.
(D)

$\Delta H_2 = (C_{calorimeter}+ C_{product})\times (T_2-T_1)$ .

Hence,option A,B and C is correct.

Find the

$\Delta H$ of the following reaction.

$OF_2(g) + H_2O(g)\longrightarrow O_2(g) + 2HF(g)$ . Average bond energies of

$O-F$ ,

$O-H$ ,

$O=O$ and

$H-F$ are

$44$ ,

$111$ ,

$118$ and

$135$ kcal mol

$^{ -1 }$ , respectively.

0%
$\Delta H= -78\ kcal$
0%
$\Delta H= +78\ kcal$
0%
$\Delta H= -498\ kcal$
0%
$\Delta H= +498 \ kcal$

Explanation

As we know,

$\Delta H=$ Bond energy data of formation of bond + Bond energy data of dissociation of bond

So for

$OF_2 (g)+H_2O(g)⟶O_2(g)+2HF(g)$ ,

$\Delta H_{reac} = \Delta H_{BE reactant}- \Delta H_{BEproduct}$

$\Delta H = 2\times 44+2\times 111 - 118 - 2\times 135 = -78\ kcal/mol$

Option A is correct.

The thermo-chemical equation for solid and liquid rocket fuel are given below:

$2Al(s) + 1\frac { 1 }{ 2 }O_2(g)\longrightarrow Al_2O_3(s), \Delta H = -1667.8 kJ$

$H_2(g) +\frac { 1 }{ 2 }O_2(g)\longrightarrow H_2O(l), \Delta H = -285.9 kJ$ ,

Then,

$\Delta H$ for the reaction :

$Al_2O_3(s)\longrightarrow 2Al(s) + 1\frac { 1 }{ 2 }O_2(g)$ is :

0%
$\ 1667.8\ kJ mol^{ -1 }$
0%
$\ 1222.3\ kJ mol^{ -1 }$
0%
$\ 1458.2\ kJ mol^{ -1 }$
0%
$\ 1785.3\ kJ mol^{ -1 }$

Explanation

As given,

$2Al(s) + 1\frac { 1 }{ 2 }O_2(g)\longrightarrow Al_2O_3(s), \Delta H = -1667.8 kJ$

and by this we can see that, for reverse reaction :

$Al_2O_3(s)\longrightarrow 2Al(s) + 1\frac { 1 }{ 2 }O_2(g); \Delta H = 1667.8 kJ$

Which of the following has the same value as

$\Delta_f H^{ \ominus }$ ,

$CO$
a.

$\frac { 1 }{ 2 } \Delta_f H^{ \ominus } (CO_2)$
b.

$\frac { 1 }{ 2 } \Delta_c H^{ \ominus } (graphite)$
c.

$\Delta_f H^{ \ominus } (CO_2) - \Delta_f H^{ \ominus } (graphite)$
d.

$\Delta_c H^{ \ominus } (graphite) - \Delta _c H^{ \ominus } (CO)?$

0%
a
0%
b
0%
c
0%
d

Explanation

As we know,

$C(graphite) + O_2(g) \rightarrow CO_2(g) ;\quad \Delta _cH =x\ kJ$ ................1

$CO + 1/2 O_2(g) \rightarrow CO_2(g)$ ;

$\Delta_c H^{ \ominus }$

$(CO) = z\ kJ$ .......................2

Equation (1)-(2) gives (3) eqaution.

$C(graphite) + 1/2 O_2(g) \rightarrow CO(g)$ ;

$\Delta_f H^{ \ominus }{(CO)}=y\ kJ$ .......................3

From these equations, it can be seen that

$\Delta_f H^{ \ominus }$

$(CO)$ =

$\Delta_c H^{ \ominus } (graphite) - \Delta _c H^{ \ominus } (CO)$

If the resonance energy of

$NO_2(:O--N==O:)$ is X kJ The measured enthalpy formation of

$NO_2(\Delta_f H^{ \ominus } )$ is

$34 kJ mol^{ -1 }$ . The bond energies given are:

$N--O \Rightarrow 222 kJ mol^{ -1 }$

$N\equiv N \Rightarrow 946 kJ mol^{ -1 }$

$O==O \Rightarrow 498 kJ mol^{ -1 }$

$N==O \Rightarrow 607 kJ mol^{ -1 }$
Find out the value of X

0%
$X= -108$
0%
$X= +108$
0%
$X= -59$
0%
None of these

Explanation

Sol.

$R.E. = \Delta_f H(observed) - \Delta_f H(calculated)$
The balanced chemical reaction for the formation of nitrogen dioxide is as shown.

$\frac { 1 }{ 2 } N_2(g) + O_2(g)\longrightarrow NO_2(g)$

$\Delta_f H_{ cal } = \text {(BE of reactants - BE of products)}$

$=\begin{pmatrix} \frac { 1 }{ 2 } \text {BE of } N_2 +\text {BE of } O_2\end{pmatrix} - \begin{pmatrix} \frac { 1 }{ 2 } \text {BE of } N==O + \text {BE of } N--O\end{pmatrix}$

$=\begin{pmatrix} \frac { 1 }{ 2 }\times 946 + 498\end{pmatrix} - (607 + 222)$

$= 971-829$

$=142 kJ$

$\Delta H_{(observed)}$

$= 34 kJ$

$\therefore$ Resonacne .Energy

$= 34-142$

$= -108 kJ$
Hence, the value of

$X$ is

$-108$ .

A change in the free energy of a system at constant temperature and pressure will be:

$\Delta_{ sys }G = \Delta_{ sys }H - T\Delta_{ sys }S$

At constant temperature and pressure,

$\Delta_{ sys }G < 0 (spontaneous)$
$\Delta_{ sys }G = 0 (equilibrium)$
$\Delta_{ sys }G > 0 (non-spontaneous)$

For a system in equilibrium,

$\Delta G = 0$ , under conditions of constant_____

0%
temperature and pressure
0%
pressure and volume
0%
temperature and volume
0%
energy and volume

Calculate the

$\Delta H^{ \ominus }$ for the reduction of

$Fe_2O_3(s)$ by

$Al(s)$ at

$25^{ \circ }C$ . The enthalpies of formation of

$Fe_2O_3(s)$ and

$Al_2O_3$ are

$-825.5$ and

$-1675.7 kJ mol^{ -1 }$ respectively.

0%
$\Delta H= - 850.2 kJ mol^{ -1 }$
0%
$\Delta H= +850.2 kJ mol^{ -1 }$
0%
$\Delta H= - 2500 kJ mol^{ -1 }$
0%
None of these

Explanation

The reduction of

$Fe_2O_3$ by Al is represented by following chemical reaction:

$Fe_2O_3(s) + 2Al(s)\longrightarrow 2Fe(s) + Al_2O_3; \ \ \ \ \ \ \Delta H = ?$ ......(1)

The balanced chemical reaction for the formation of

$Fe_2O_3$ is as shown below.

$3Fe(s) + \cfrac { 3 }{ 2 }O_2(g)\longrightarrow Fe_2O_3; \ \ \ \ \ \ \ \ \ \Delta H_1 = -825.5 kJ mol^{ -1 }$ ......(ii)

The balanced chemical reaction for the formation of

$Al_2O_3$ is as shown below.

$2Al(s) +\cfrac { 3 }{ 2 }O_2(g)\longrightarrow Al_2O_3; \ \ \ \ \ \ \ \ \Delta H_2 = -1675.7 kJ mol^{ -1 }$ ......(iii)

Subtract the enthalpy change for the third reaction from the enthalpy change for the second reaction to obtain the enthalpy change for the first reaction.

$\Delta H = \Delta H_1 - \Delta H_2$

$= -1675.7 - (-825.5)$

$= - 850.2 kJ mol^{ -1 }$

Hence, option A is correct.

4 grams of sodium hydroxide pellets were dissolved in

$100 {cm}^{3}$ of water. The temperature before adding the sodium hydroxide pellets was 25 degrees C, and after adding the pellets it was 35 degrees C. Calculate the enthalpy change in

${kJ}/{mole}$ of the reaction.

[Specific heat capacity of water

$= 4.2 {J}/{k}/{g}]$

0%
$42 {kJ}/{mole}$
0%
$4.2 {kJ}/{mole}$
0%
$4200 {kJ}/{mole}$
0%
None

Explanation

Heat evolve

$= m{C}_{v}\Delta t = 100 \times 4.2 \times 10 = 4.2 kJ$
for 0.1 mole the enthalpy change

$= 4.2 kJ$
for 1 mole the enthalpy change

$= 42 kJ$

How much heat is liberated when one mole of gaseous

$Na^{ \oplus }$ combines with one mole of

$Cl^{ \ominus }$ ion to form solid

$NaCl$ .

Use the data given below:

$Na(s) + \frac { 1 }{ 2 } Cl_2(g) \longrightarrow NaCl(s)$ ;

$\Delta H = -98.23 kcal$

$Na(s)\longrightarrow Na(g)$ ;

$\,\,\,\, \Delta H = +25.98 kcal$

$Na(g)\longrightarrow Na^{ \oplus } + e^ { - }$ ;

$\,\,\,\,\,\,\,\Delta H = +120.0 kcal$

$Cl_2(g)\longrightarrow 2Cl(g)$ ;

$\,\,\,\,\,\,\,\,\Delta H = +58.02 kcal$

$Cl^{ \ominus } (g)\longrightarrow Cl(g) + e^{ - }$ ;

$\,\,\,\,\,\,\,\Delta H = +87.3 kcal$

0%
$\Delta H = -185.92 Kcal$
0%
$\Delta H = +185.92 Kcal$
0%
$\Delta H = 0 Kcal$
0%
None of these

Explanation

$Na^{ \oplus } (g) + Cl^{ \ominus } (g)\longrightarrow NaCl(s)$

$;\,\,\,\,\,\,\Delta H = ?$

Given:

$Na(s) + \frac { 1 }{ 2 } Cl_2(g)\longrightarrow NaCl(s)$

$;\,\,\,\,\,\,\,\Delta H = -98.23 kcal$

$Na(s)\longrightarrow Na(g)$

$;\,\,\,\,\,\,\,\,\Delta H = 25.98 kcal$

$Na(g)\longrightarrow Na^{ \oplus } (g) + e^{ - }$

$;\,\,\,\,\,\,\,\,\Delta H = 120.0 kcal$

$\frac { 1 }{ 2 } Cl_2(g)\longrightarrow Cl(g)$

$;\,\,\,\,\,\,\,\,\Delta H = 58.02\times \frac { 1 }{ 2 } kcal$

$Cl^{ \ominus} (g)\longrightarrow Cl(g) + e^{ - }$

$;\,\,\,\,\,\,\,\,\,\Delta H =87.3 kcal$

Rewriting equations:

$Na(s) + \frac { 1 }{ 2 } Cl_2(g)\longrightarrow NaCl(s)$

$;\,\,\,\,\,\,\,\,\Delta H_1 = -98.23$

$Na(g)\longrightarrow Na(s)$

$;\,\,\,\,\,\,\,\,\Delta H_2 = -25.98$

$Na^{ \oplus } (g) + e^{ - }\longrightarrow Na(g)$

$;\,\,\,\,\,\,\,\,\Delta H_3 = -120.0$

$\frac { 1 }{ 2 } [2Cl (g)\longrightarrow Cl_2 (g) ]$

$;\,\,\,\,\,\,\,\,\Delta H_4 = -58.02\times \frac { 1 }{ 2 }$

$Cl^{ \ominus} (g)\longrightarrow Cl(g) + e^{ - }$

$;\,\,\,\,\,\,\,\,\,\Delta H_5 = 87.3$

$Na^{ \oplus }(g) + Cl^{ \ominus }\longrightarrow NaCl(s)$

$\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 + \Delta H_4 + \Delta H_5$

$\,\,\,\,\,\,\,\,\ = -98.23 - 25.98 - 120.0 - \frac{ 1 }{ 2 }\times 58.02 + 87.3$

$\,\,\,\,\,\,\, \ = -185.92 Kcal$

$\Delta_f H^{ \ominus }$ of Cyclohexene

$(l)$ and benzene at

$25^{ \circ}C$ is

$-156$ and

$+46 kJ mol^{ -1 }$ , respectively.

$\Delta_{ hydrogenation }H^{ \ominus }$ of Cyclohexene

$(l)$ at

$25^{ \circ }C$ is

$-119 kJ mol^{ -1 }$ .

Resonance energy of benzene is found to be

$-38x kJ mol^{ -1 }$ . Find the value of x.

0%
4
0%
2
0%
8
0%
None of these

Explanation

This shows that the generation of one

$(C=C)$ bond in Cychlohexene requires

$119 kJ mol^{ -1 }$ of enthalpy. TO calculate RE resonance energy.

$\Delta H^{ \ominus }_1 = \Delta H^{ \ominus }_2 = \Delta H^{ \ominus }_3 = 119 kJ mol^{ -1 }, \Delta H^{ \ominus }_4 = RE$

$\Delta H^{ \ominus }_5 = \Delta_f H^{ \ominus }(benzene) - \Delta_f H^{ \ominus }(cyclohexene)$

$= 46 - (-156) = 205 kJ mol^{ -1 }$

$\therefore\quad From\quad Hess`\quad law:$

$\Delta H^{ \ominus }_4 = \Delta H^{ \ominus }_5 - (\Delta H^{ \ominus }_1 + \Delta H^{ \ominus }_2 + \Delta H^{ \ominus } _3)$ .

$= 205 - 3\times 119 = -152 kJ mol^{ -1 }$

$\therefore -38x = -152$

$x = 4$

$2Al(s) + 1\frac { 1 }{ 2 }O_2(g)\longrightarrow Al_2O_3(s), \Delta H = -166.78 kJ$

$H_2(g) + \frac { 1 }{ 2 }O_2(g)\longrightarrow H_2O(l), \Delta H = -285.9 kJ$

Then

$\Delta H$ for the reaction,

$Al_2O_3(s)\longrightarrow 2Al(s) + 1\frac { 1 }{ 2 }O_2(g)$ is:

0%
$\Delta H = 174.20kJ mol^{ -1 }$
0%
$\Delta H = 124.55 kJ mol^{ -1 }$
0%
$\Delta H = 166.78 kJ mol^{ -1 }$
0%
$\Delta H = 156.33 kJ mol^{ -1 }$

Explanation

As given,

$2Al(s) + 1\frac { 1 }{ 2 }O_2(g)\longrightarrow Al_2O_3(s), \Delta H = -166.78 kJ$

and by this we can see that,

for reverse reaction

$Al_2O_3(s)\longrightarrow 2Al(s) + 1\frac { 1 }{ 2 }O_2(g) \Delta H = 166.78 kJ$

Determine

${ \Delta H }/{ kJ }$ for the following reaction using the listed enthalpies of reaction:

$4CO\left( g \right) +8{ H }_{ 2 }\left( g \right) \longrightarrow 3C{ H }_{ 4 }\left( g \right) +C{ O }_{ 2 }\left( g \right) +2{ H }_{ 2 }O\left( l \right)$

$C\left( graphite \right) +{ 1 }/{ 2 }{ O }_{ 2 }\left( g \right) \longrightarrow CO\left( g \right) ; { \Delta H }/{ kJ }=-110.5kJ$

$CO\left( g \right) +{ 1 }/{ 2 }{ O }_{ 2 }\left( g \right) \longrightarrow C{ O }_{ 2 }\left( g \right) ; { \Delta H }/{ kJ }=-282.9kJ$

${ H }_{ 2 }\left( g \right) +{ 1 }/{ 2 }{ O }_{ 2 }\left( g \right) \longrightarrow { H }_{ 2 }O\left( l \right) ; { \Delta H }/{ kJ }=-285.8kJ$

$C\left( graphite \right) +2{ H }_{ 2 }\left( g \right) \longrightarrow C{ H }_{ 4 }\left( g \right) ; { \Delta H }/{ kJ }=-74.8kJ$

0%
-622.4
0%
-686.2
0%
-747.5
0%
-653.5

Explanation

$4CO\left( g \right) +8{ H }_{ 2 }\left( g \right) \longrightarrow 3C{ H }_{ 4 }\left( g \right) +C{ O }_{ 2 }\left( g \right) +2{ H }_{ 2 }O\left( l \right)$

$C\left( graphite \right) +{ 1 }/{ 2 }{ O }_{ 2 }\left( g \right) \longrightarrow CO\left( g \right) ; { \Delta H }/{ kJ }=-110.5kJ$ ---1

$CO\left( g \right) +{ 1 }/{ 2 }{ O }_{ 2 }\left( g \right) \longrightarrow C{ O }_{ 2 }\left( g \right) ; { \Delta H }/{ kJ }=-282.9kJ$ ----2

${ H }_{ 2 }\left( g \right) +{ 1 }/{ 2 }{ O }_{ 2 }\left( g \right) \longrightarrow { H }_{ 2 }O\left( l \right) ; { \Delta H }/{ kJ }=-285.8kJ$ ----- 3

$C\left( graphite \right) +2{ H }_{ 2 }\left( g \right) \longrightarrow C{ H }_{ 4 }\left( g \right) ; { \Delta H }/{ kJ }=-74.8kJ$ ----- 4

Applying Hess's law,

$\Delta { H }_{ r } =3\times (-\Delta H_1)+\Delta H_2+2\times \Delta H_3+3\times\Delta H_4$

$\Delta { H }_{ r } = \left[ 3\left( 110.5 \right) - 282.9 + 2\left( -285.8 \right) + 3\left( -74.8 \right) \right]$

$\Delta { H }_{ r } = -747.4\ kJ$

$\Delta_f H^{ \ominus }$ of hypothetical

$MgCl$ is

$-125 kJ mol^{ -1 }$ and for

$MgCl_2$ is

$-642 kJ mol^{ -1 }$ . The enthalpy of disproportionation of

$MgCl$ is

$-49x$ . Find the value of x.

0%
$x = 8$
0%
$x = 8.5$
0%
$x = 16$
0%
None of these

Explanation

The reaction for formation of MgCl is as shown.

$Mg(s) + 1/2Cl_2(g)\longrightarrow MgCl$ ......(i)

$\Delta H_1 = -125 kJ mol^{ -1 }$
The reaction for formation of

$MgCl_2$ is as shown

$Mg(s) + Cl_2(g)\longrightarrow MgCl_2$ ......(ii)

$\Delta H_2 = -642 kJ mol^{ -1 }$
The disproportionation reaction is as shown.

$2MgCl\longrightarrow Mg + MgCl_2$ ......(iii)

$\Delta H = ?$
The reaction (i) is multiplied by 2 and subtracted from reaction (ii) to obtain reaction (iii)
The enthalpy of disproportionation of MgCl is

$\Delta H = \Delta H_2 - 2\Delta H_1 = -642 - (2\times -125) = -392 kJ mol^{ -1 }$

$\therefore -49x = -392$

$x = 8$

Using the enthalpies of formation, calculate the energy (kJ) released when

$3.00 g$ of

$N{ H }_{ 3\left( g \right) }$ reacts according to the following equation?
(Atomic weights:

$N = 14.00, H = 1.008$ ).

$4N{ H }_{ 3 }\left( g \right) +5{ O }_{ 2 }\left( g \right) \longrightarrow 4NO\left( g \right) +6{ H }_{ 2 }O\left( g \right)$

$\Delta HN{ H }_{ 3 }\left( g \right) =-46.1 { kJ }/{ mole }$

$\Delta HNO\left( g \right) =+90.2 { kJ }/{ mole }$

$\Delta H{ H }_{ 2 }O\left( g \right) =-241.8 { kJ }/{ mole }$

0%
$34.3$
0%
$30.8$
0%
$39.9$
0%
$42.6$

Explanation

Energy released, when four mole of

$NH_{ 3 }$ reacts, is given by

$\displaystyle \Delta_{r}H = H_{ Products } - H_{ Reactants }$

$\displaystyle \Delta_{r}H = (4 \times \Delta HNO\left( g \right) + 6 \times \Delta H{ H }_{ 2 }O\left( g \right)) - (4 \times \Delta HN{ H }_{ 3 }\left( g \right) + 5 \times \Delta H{ O }_{ 2 }\left( g \right))$

$\displaystyle \Delta_{r}H = 4 \times 90.2 + 6 \times (-241.8) - 4 \times (-46.1) + 5 \times 0$ (As

$\Delta H{ O }_{ 2 }\left( g \right) = 0$ )

$\displaystyle \Delta_{r}H = -905.6\ kJ$ Thus, for four mole of

$NH_{ 3 }$ , i.e.,

$(4\times14=) 68g$ of

$NH_{ 3 }$ amount of energy released is 905.6 kJ. Therefore, for 3g of

$NH_{ 3 }$ amount of energy released will be

$\dfrac{(905.16 \times 3)}{68} = 39.9\ kJ$

Calculate

${ \Delta H }/{ kJ }$ for the following reaction using the listed standard enthalpy of reaction data.

$2{ N }_{ 2 }\left( g \right) +5{ O }_{ 2 }\left( g \right) \longrightarrow 2{ N }_{ 2 }{ O }_{ 5 }\left( s \right)$

${ N }_{ 2 }\left( g \right) +3{ O }_{ 2 }\left( g \right) +{ H }_{ 2 }\left( g \right) \longrightarrow 2HN{ O }_{ 3 }\left( aq \right) ; \ { \Delta H }/{ kJ }=-414.0$

${ N }_{ 2 }{ O }_{ 5 }\left( s \right) +{ H }_{ 2 }O\left( l \right) \longrightarrow 2HN{ O }_{ 3 }\left( aq \right) ; \ { \Delta H }/{ kJ }=-86.0$

$2{ H }_{ 2 }\left( g \right) +{ O }_{ 2 }\left( g \right) \longrightarrow 2{ H }_{ 2 }O\left( l \right) ; \ { \Delta H }/{ kJ }=-571.6$

0%
-84.4
0%
-243.6
0%
-71.2
0%
-121.8

Explanation

$2{ N }_{ 2 }\left( g \right) +5{ O }_{ 2 }\left( g \right) \longrightarrow 2{ N }_{ 2 }{ O }_{ 5 }\left( s \right)$ ----- a

Given that,

${ N }_{ 2 }\left( g \right) +3{ O }_{ 2 }\left( g \right) +{ H }_{ 2 }\left( g \right) \longrightarrow 2HN{ O }_{ 3 }\left( aq \right) ; { \Delta H }/{ kJ }=-414.0$ -------- 1

${ N }_{ 2 }{ O }_{ 5 }\left( s \right) +{ H }_{ 2 }O\left( l \right) \longrightarrow 2HN{ O }_{ 3 }\left( aq \right) ; { \Delta H }/{ kJ }=-86.0$ -------- 2

$2{ H }_{ 2 }\left( g \right) +{ O }_{ 2 }\left( g \right) \longrightarrow 2{ H }_{ 2 }O\left( l \right) ; { \Delta H }/{ kJ }=-571.6$ ------------- 3

Applying Hess's law,
Eqn (a) = 2 x Eqn (1) - 2 x Eqn (2) - Eqn (3)

$\Delta { H }_{ r }=\left[ 2\left( -414 \right) +2\left( 86 \right) +571.6 \right]$

$\Delta { H }_{ r }=-84.4\ kJ$

Calculate the value of

${ \Delta H }{\ ( kJ) }$ for the following reaction using the listed thermochemical equations

$2C(s) + H_2(g) \rightarrow C_2H_ 2 (g)$ .

$2{ C }_{ 2 }{ H }_{ 2 }\left( g \right) +5{ O }_{ 2 }\left( g \right) \longrightarrow 4C{ O }_{ 2 }\left( g \right) +2{ H }_{ 2 }O\left( l \right)$ ;

$\Delta H^o= -2600\ kJ$

$C\left( s \right) +{ O }_{ 2 }\left( g \right) \longrightarrow C{ O }_{ 2 }\left( g \right)$ ;

$\Delta H^o=-390\ kJ$

$2{ H }_{ 2 }\left( g \right) +{ O }_{ 2 }\left( g \right) \longrightarrow 2{ H }_{ 2 }O\left( l \right)$ ;

$\Delta H^o =-572\ kJ$

0%
$+184$
0%
$+214$
0%
$+202$
0%
$+234$

Use the given standard enthalpies of formation to determine the heat of reaction of the following reaction:

${ C }_{ 2 }{ H }_{ 5 }OH\left( l\right) +3{ O }_{ 2 }\left( g \right) \longrightarrow 2C{ O }_{ 2 }\left( g \right) +3{ H }_{ 2 }O\left( g \right)$

$\Delta { H }_{ f }^{ }{ C }_{ 2 }{ H }_{ 5 }OH\left( l \right) =-277.7{ kJ }/{ mole }$

$\Delta { H }_{ f }^{ }C{ O }_{ 2 }\left( g \right) =-393.5{ kJ }/{ mole }$

$\Delta { H }_{ f }^{ }{ H }_{ 2 }O\left( g \right) =-241.8{ kJ }/{ mole }$

0%
-1456.3
0%
-1234.7
0%
-1034.0
0%
-1119.4

Explanation

$\Delta { H }_{ r }=\left[ 2{ \left( \Delta { H }_{ f } \right) }_{ C{ O }_{ 2 } }+3{ \left( \Delta { H }_{ f } \right) }_{ { H }_{ 2 }O }-{ \left( \Delta { H }_{ f } \right) }_{ { C }_{ 2 }{ H }_{ 5 }OH } \right]$

$\Delta { H }_{ r }=\left[ 2\left( -393.5 \right) -3\left( 241.8 \right) +277.7 \right]$

$\Delta { H }_{ r }=-1234.7{ kJ }/{ mole }$

The enthalpy change for the following process at

${25}^{o}C$ and under constant pressure at

$1$ atm are as follows:

${CH}_{4}(g)\longrightarrow C(g)+4H(g)$

${\Delta}_{r}H=396kcal/mole$

${C}_{2}{H}_{6}(g)\longrightarrow 2C(g)+6H(g)$

${\Delta}_{r}H=676kcal/mole$

Calculate

$C-C$ bond energy in

${C}_{2}{H}_{6}$ and heat formation of

${C}_{2}{H}_{6}(g)$ .

[Given:

${\Delta}_{sub}C(s)=171.8kcal/mole$

$B.E(H-H)=104.1kcal/mole$ ]

0%
$B.E(C-C)=82kcal/mol, {\Delta}_{f}H[{C}_{2}{H}_{6}(g)]=-20.1kcal/mol$
0%
$B.E(C-C)=-82kcal/mol, {\Delta}_{f}H[{C}_{2}{H}_{6}(g)]=-20.1kcal/mol$
0%
$B.E(C-C)=82kcal/mol, {\Delta}_{f}H[{C}_{2}{H}_{6}(g)]=+20.1kcal/mol$
0%
None of these

Explanation

$\displaystyle {\Delta}_{f}H[{C}_{2}{H}_{6}(g)]=-676+343.6+312.3=-676+655.9=-20.1kcal/mol$

$\displaystyle 4(B.E(C-H))=396 (B.E(C-C))+6(99)=676$

$\displaystyle B.E(C-C)=99\ (C-C)=676-594=82kcal/mol$

Use the given standard enthalpies of formation to determine the heat of reaction of the following reaction:

$2LiOH+C{ O }_{ 2 }\left( g \right) \longrightarrow { Li }_{ 2 }C{ O }_{ 3 }\left( s \right) +{ H }_{ 2 }O\ (l)$

$\Delta { H }_{ f }^{ }LiOH\left( s \right) =-487.23{ kJ }/{ mole }$

$\Delta { H }_{ f }^{ }{ Li }_{ 2 }C{ O }_{ 3 }\left( s \right) =-1215.6{ kJ }/{ mole }$

$\Delta { H }_{ f }^{ }{ H }_{ 2 }O\ (l) =-285.85{ kJ }/{ mole }$

$\Delta { H }_{ f }^{ }C{ O }_{ 2 }\left( g \right) =-393.5{ kJ }/{ mole }$

0%
$+303.4$
0%
$-133.5$
0%
$-198.6$
0%
$+198.6$

Explanation

$2LiOH+C{ O }_{ 2 }\left( g \right) \longrightarrow { Li }_{ 2 }C{ O }_{ 3 }\left( s \right) +{ H }_{ 2 }O\left( l\right)$

$\Delta { H }_{ f }^{ }LiOH\left( s \right) =-487.23\;{ kJ }/{ mole }$

$\Delta { H }_{ f }^{ }{ Li }_{ 2 }C{ O }_{ 3 }\left( s \right) =-1215.6\;{ kJ }/{ mole }$

$\Delta { H }_{ f }^{ }{ H }_{ 2 }O\left( l \right) =-285.85\;{ kJ }/{ mole }$

$\Delta { H }_{ f }^{ }C{ O }_{ 2 }\left( g \right) =-393.5\;{ kJ }/{ mole }$

As we know,

Heat of reaction = Heat of formation of products - Heat of formation of reactants

$={(-1215.6-285.85} )-\ {((2\times-487.23) +(-393.5))} = -133.5$ kJ/mole

Determine

$\Delta H$ of the following reaction using the listed heats of formation:

$4HN{ O }_{ 3 }\left( I \right) +{ P }_{ 4 }{ O }_{ 10 }\left( s \right) \longrightarrow 2{ N }_{ 2 }{ O }_{ 5 }\left( s \right) +4HP{ O }_{ 3 }\left( s \right)$

$\Delta { H }_{ f }^{ }HN{ O }_{ 3 }\left( I \right) =-174.1{ kJ }/{ mole }$

$\Delta { H }_{ f }^{ }{ N }_{ 2 }{ O }_{ 5 }\left( s \right) =-43.1{ kJ }/{ mole }$

$\Delta { H }_{ f }^{ }{ P }_{ 4 }{ O }_{ 10 }\left( s \right) =-2984.0{ kJ }/{ mole }$

$\Delta { H }_{ f }^{ }HP{ O }_{ 3 }\left( s \right) =-948.5{ kJ }/mole$

0%
-176.3
0%
-199.8
0%
+276.2
0%
-242.4

Explanation

The enthalpy change for the reaction is the difference between the enthalpy of products and the enthalpy of reactants.

$\Delta { H }_{ r }=\left[ 2{ \left( \Delta { H }_{ f } \right) }_{ { N }_{ 2 }{ O }_{ 5 } }+4{ \left( \Delta { H }_{ f } \right) }_{ HP{ O }_{ 3 } }-4{ \left( \Delta { H }_{ 5 } \right) }_{ HN{ O }_{ 3 } }-{ \left( \Delta { H }_{ f } \right) }_{ { P }_{ 4 }{ O }_{ 10 } } \right]$

$\Delta { H }_{ r }=\left[ 2\left( -43.1 \right) +4\left( -948.5 \right) -4\left( -174.1 \right) -\left( -2984.0 \right) \right]$

$=-199.8 \: kJ/mol$
Hence, the enthalpy change for the reaction is -199.8 kJ/mol.

Caesium chloride is formed according to the following equation:

$Cs(s)+0.5{Cl}_{2}(g)\longrightarrow CsCl(s)$

The enthalpy of sublimation of

$Cs$ , enthalpy of dissociation of chlorine, ionization energy of

$Cs$ and electron affinity of chlorine are

$81.2, 243.0, 375.7$ and

$-348.3kJ$

${ol}^{-1}$ . The energy change involved in the formation of

$CsCl$ is

$388.6\ kJ.{mol}^{-1}$ . Calculate the lattice energy of

$CsCl$ .

0%
$-618.7\ kJ{mol}^{-1}$
0%
$+618.7\ kJ{mol}^{-1}$
0%
$1315.2\ kJ{mol}^{-1}$
0%
None of these

Determine

${ \Delta H }/{ kJ }$ for the following reaction using the listed enthalpies of reaction:

$4CO\left( g \right) +8{ H }_{ 2 }\left( g \right) \longrightarrow 3C{ H }_{ 4 }\left( g \right) +C{ O }_{ 2 }\left( g \right) +2{ H }_{ 2 }O\left( l \right)$

Given that

$C\left( graphite \right) +{ 1 }/{ 2 }{ O }_{ 2 }\left( g \right) \longrightarrow CO\left( g \right) ; \ { \Delta H }/{ kJ }=-110.5kJ$

$CO\left( g \right) +{ 1 }/{ 2 }{ O }_{ 2 }\left( g \right) \longrightarrow C{ O }_{ 2 }\left( g \right) ;\ { \Delta H }/{ kJ }=-282.9kJ$

${ H }_{ 2 }\left( g \right) +{ 1 }/{ 2 }{ O }_{ 2 }\left( g \right) \longrightarrow { H }_{ 2 }O\left( l \right) ;\ { \Delta H }/{ kJ }=-285.8kJ$

$C\left( graphite \right) +2{ H }_{ 2 }\left( g \right) \longrightarrow C{ H }_{ 4 }\left( g \right) ;\ { \Delta H }/{ kJ }=-74.8kJ$

0%
$584.9\:kJ\:mol^{-1}$
0%
$279.8\:kJ\:mol^{-1}$
0%
$747.4\:kJ\:mol^{-1}$
0%
$925\:kJ\:mol^{-1}$

Explanation

$4CO\left( g \right) +8{ H }_{ 2 }\left( g \right) \longrightarrow 3C{ H }_{ 4 }\left( g \right) +C{ O }_{ 2 }\left( g \right) +2{ H }_{ 2 }O\left( l \right) ;$

$\Delta H=?$

$C\left( graphite \right) +{ 1 }/{ 2 }{ O }_{ 2 }\left( g \right) \longrightarrow CO\left( g \right) ; \ { \Delta H_1 }=-110.5kJ$

$CO\left( g \right) +{ 1 }/{ 2 }{ O }_{ 2 }\left( g \right) \longrightarrow C{ O }_{ 2 }\left( g \right) ;\ { \Delta H_2 }=-282.9kJ$

${ H }_{ 2 }\left( g \right) +{ 1 }/{ 2 }{ O }_{ 2 }\left( g \right) \longrightarrow { H }_{ 2 }O\left( I \right) ;\ { \Delta H_3 }=-285.8kJ$

$C\left( graphite \right) +2{ H }_{ 2 }\left( g \right) \longrightarrow C{ H }_{ 4 }\left( g \right) ;\ { \Delta H_4 }=-74.8kJ$

Applying Hess's law,

$\Delta H_r =-3\times \Delta H_1+\Delta H_2+2\times \Delta H_3+ 3\times \Delta H_4$

$\Delta { H }_{ r } = \left[ 3\left( 110.5 \right) - 282.9 + 2\left( -285.8 \right) + 3\left( -74.8 \right) \right]$

$= -747.4$

Using the data (all values are in

$kJ/mol$ at

${25}^{o}C$ ) given below:

(i) Enthalpy of polymerization of ethylene

$=-72$
(ii) Enthalpy of formation of benzene

$(l)=49$
(iii) Enthalpy of vaporization of benzene

$(l)=30$
(iv) Resonance energy of benzene

$(l)=-152$
(v) Heat of formation of gaseous atoms from the elements in their standard states

$H=218, C=715$ .

Average bond energy of

$C-H=415$ . Calculate the

$B.E$ of

$C-C$ and

$C=C$ .

0%
$C-C=343.67$ , $C=C=615.33$
0%
$C-C=615.33$ , $C=C=342.67$
0%
$C-C=343.67$ , $C=C=959$
0%
None of these

Explanation

Given
For the polymerization of ethylene,

$n({CH}_{2}={CH}_{2})\rightarrow {(-{CH}_{2}-{CH}_{2}-)}_{n};$

$\Delta H=-72$

i.e.,

${B}_{C=C}-2{B}_{C-C}=-72\ \ .......(i)$

For the formation of benzene

$6{C}_{(s)}+3{H}_{2(g)}\longrightarrow {C}_{6}{H}_{6(l)};$

$\Delta H=49\ \ ......(ii)$

For the vaporisation of benzene

${C}_{6}{H}_{6(l)}\rightarrow {C}_{6}{H}_{6(g)}$

$\Delta H=30$

R.E of

${C}_{6}{H}_{6}=-152$

For the formation of H atom from hydrogen molecule in standard state,

$\cfrac{1}{2} {H}_{2}\rightarrow H;$

$\Delta H=218$

For the formation of gaseous C atom

${C}_{(s)}\rightarrow {C}_{(g)};$

$\Delta H=715$

The average bond energy

${B}_{C-H}=415$ for equation (2)

$(6\times 715+6\times 218)-(3{B}_{C-C}+3{B}_{C=C}+6\times 415-(-152))=49+30$

${B}_{C-C}+{B}_{C=C}=959..........(iii)$

from equation

$(i)$ and

$(iii)$

The bond energy of

$C-C$ bond

${B}_{C-C}=343.66$ and the bond energy of C=C bond

${B}_{C}=C=615.33$

Use the given standard enthalpies of formation to determine the heat of reaction of the following reaction:

$TiCL_4(g)+2\:H_2O(g)\rightarrow TiO_2(g)+4\:HCl(g)$

$\Delta H^{\circ}_f\:TiCL_4(g)=-763.2\:kJ/mol$

$\Delta H^{\circ}_f\:TiO_2(g)=-944.7\:kJ/mol$

$\Delta H^{\circ}_f\:H_2O(g)=-241.8\:kJ/mol$

$\Delta H^{\circ}_f\:HCl(g)=-92.3\:kJ/mol$

0%
$-\:278.1$
0%
$+\:369.2$
0%
$+\:67.1$
0%
$-\:67.1$

Explanation

The enthalpy change for the reaction is the difference in the enthalpies of formation of products and the enthalpies of formation of reactants.

$\Delta H_r=[(\Delta H_f)_{TiO_2}+4(\Delta H_f)_{HCl} - [(\Delta H_f)_{TiCl_4}+2(\Delta H_f)_{H_2O}]$

Substitute values in the above expression.

$\Delta H_r=-944.7 +(4\times -92.3) - [-763.2+(2\times -241.8)$

$\Delta H_r=-67.1\:kJ/mol$

Hence, the heat of reaction for the hydrolysis of titanium tetrachloride is - 67.1 kJ/mol.

How many joules of heat are absorbed when 70.0 g of water is completely vaporised at its boiling point ?
[Take : LV = 2260 kJ / kg]

0%
22352
0%
52460
0%
22344
0%
158200

Anhydrous

$AlCl_3$ is a covalent compound. From the data given below, predict whether it would remain covalent or become ionic in an aqueous solution .

Ionisation energy of

$Al = 5137 \ kJ mol^{-1}$ ,

$\Delta H$ hydration for

$Al^{3+}=-4665\:kJ\:mol^{-1};$

$\Delta H$ hydration for

$Cl^-=-381\:kJ\:mol^{-1}$

0%
Ionic
0%
Covalent
0%
Partially ionic
0%
Partially covalent

Explanation

Total hydration energy is

$-4665+ 3\times (-381)=-5808\ kJ mol^{-1}$ and ionization energy is

$5137\ kJ mol^{-1}$ . So some energy will use in breaking bond. So,

$AlCl_3$ become ionic in aqueous solution.

A piston filled with 0.04 mole of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37.0°C. As it does so, it absorbs 207 J of heat. The values of q and W for the process will be :
[Take : R = 8.314 J / mol K, In 7.5 = 2.01]

0%
q = +207, W = -207
0%
q = -207, W = -207
0%
q = -207, W = +207
0%
q = +207, W = +207

A swimmer coming out from a pool is covered with a film of water weighing about 18 g. Calculate the internal energy of vaporisation at 100°C.
[

$\Delta H_{vap}^o$ for water at 373 K = 40.66 kJ/mol ]

The correct option is:

0%
35.67 kJ
0%
36.64 kJ
0%
37.56 kJ
0%
None of the above

The enthalpies of formation of

$Al_2O_3$ and

$Cr_2O_3$ are -1596 kJ and -1134 kJ respectively.

$\Delta H$ for the reaction,

$2Al + Cr_2O_3\rightarrow 2Cr + Al_2O_3$ is :

0%
-2730 kJ
0%
-462 kJ
0%
-1365 kJ
0%
+2730 kJ

Explanation

The enthalpies of formation of Al and Cr in their standard states are considered zero.

$2Al+\frac32O_2\rightarrow Al_2O_3$

$2Cr+\frac32O_2\rightarrow Cr_2O_3$

$\displaystyle \Delta H _{rxn} = \Delta H _{Al_2O_3} - \Delta H_{Cr_2O_3}$

$\displaystyle \Delta H _{rxn} = -1596 - (-1134) = -462 \: kJ$

According to the given reaction, when a

$45$ gram sample of the ethanol is burned with excess oxygen, how much energy is released in the form of heat?

$C_2H_5OH(l) + 3O_2(g)\rightarrow 2CO_2(g) + 3H_2O(l)$

$\triangle H = -1.40\times 10^3$ kJ.

0%
0.995 kJ
0%
$5.1 \times 10^2$ kJ
0%
$1.40 \times 10^3$ kJ
0%
$2.80 \times 10^3$ kJ
0%
5000 kJ

If 10 g of liquid at 300K is heated to 350 K, the liquid absorbs 6 kcals. Determine the specific heat of the liquid in cal/

$\displaystyle { g }^{ \circ }C$ .

0%
6
0%
12
0%
60
0%
120
0%
600

Explanation

Specific heat =

$\dfrac {energy\space absorbed}{sample\space mass\times Temperature\space change} =\dfrac {6000} {10\times 50}= 12 cal/g$

Compute

$\Delta_rG$ for the reaction

$H_2O (;, 1 atm, 323 K) \rightarrow H_2O (g, 1 atm, 323 K)$
Given that :

$\Delta_{vap}H$ at

$373 K = 40.639 kJmol^{1}, C_P(H_2O, l) = 75.312 J K^{1} mol^{1}, C_P(H_2O, g) = 33.305 J K^{1}mol^{1}$

0%
$\Delta_rG = 5.59 kJ mol^{1}$
0%
$\Delta_rG = -5.59 kJ mol^{1}$
0%
$\Delta_rG = 55.9 kJ mol^{1}$
0%
None of these

Explanation

The change in the molar heat capacity at constant pressure for the reaction is as shown.

$\Delta_rC_P=C_p (H_2O,g) - C_P(H_2O, l) =33.305-75.312=-42.007 J/K mole$
The entropy change at 323 K is as shown.

$\displaystyle \Delta_rS_{323}=\frac {\Delta H}{T}=\frac {40639}{323}=108.95 J/K mole$
The relationship between the entropy change and the change in molar heat capacity at constant pressure is as shown.

$\displaystyle d(\Delta_rS)=\frac {\Delta_rC_PdT}{T}$

$\Delta_rS_{373}-\Delta_rS_{323}=\Delta_rC_P ln\frac {T_2}{T_1}$

$\Delta_rS_{373}=108.95-(-42.007 ln \frac {373}{323})$

$=115 J/K mole$
The relationship between the enthalpy change and the change in molar heat capacity at constant pressure is as shown.

$d(\Delta_rH)=\Delta_rC_PdT$

$\Delta_rH_{373} - \Delta_rH_{323} = 42.007 (50)$

$\Delta_rH_{373} = 42739.35 J/mole$
The relationship between Gibbs free energy change, the entropy change and the enthalpy change is as shown.

$\displaystyle \Delta_rG_{323}= \Delta_rH_{323} - T\Delta_rS_{323}$

$\displaystyle \Delta_rG_{323} = 42739.35 -323 (115)$

$\displaystyle \Delta_rG_{323} = 5594.35 J = 5.59 kJ/mole$

Hence, the Gibbs free energy change for the reaction is

$5.59 kJ/mol$ .

Calculate the

${ \Delta H }_{ r }$ of KH(s).

0%
62 kJ/mol
0%
124 kJ/mol
0%
31 kJ/mol
0%
None of these

Explanation

Born-Haber Cycle

It is a series of steps (chemical processes) used to calculate the lattice energy of ionic solids, which is difficult to determine experimentally. You can think of BH cycle as a special case of Hess's law which states that the overall energy change in a chemical process can be calculated by breaking down the process into several steps and adding the energy change from each step.

$({ \Delta H }_{ f })_{ KH }$ = - $\frac { 124 }{ 2 } \Rightarrow$ 62 kJ/mole.

Fixed amount of an ideal gas contained in a sealed rigid vessel

$(V = 24.6\ litre)$ at

$1.0$ bar is heated reversibly from

$27^oC$ to

$127^oC$ .

Determine change in Gibb's energy (in Joule) if entropy of gas

$S = 10 + 10^{2} T (J/K)$ .

0%
$-530$ J
0%
$+530$ J
0%
$530$ kJ
0%
None of the above

Explanation

The relationship between the Gibbs free energy, enthalpy and entropy is

$G = H - TS = U + PV - TS$

$dG = dU + PdV + VdP - TdS - SdT$

$w = 0, dV = 0, dV = dq = T dS$
Hence,

$dG = TdS + VdP - TdS - SdT$

$dG = VdP - SdT \Rightarrow \Delta G = V\Delta P-\int S dT$ ......(i)

$VdP = V (P_2 - P_1)$ ......(ii)

$\dfrac {P_2}{T_2}=\dfrac {P_1}{T_1}\Rightarrow \dfrac {P_2}{400}=\dfrac {1}{300}\Rightarrow P_2=\dfrac {4}{3} \times 1 \: bar$
Substitute value of

$P_2$ in equation (ii)

$VdP = 24.6 (4/3- 1) = 8.2 L-atm = 820 J$ ......(iii)

$\int SdT= 2 \displaystyle \int_{T_1}^{T_2} (10+0.01T)dT=10(T_2-T_1)+0.005(T_2^2-T_1^2)$

$SdT = 10(100) + 0.005 (400^2 -300^2) = 1350$ .....(iv)
Substitute (iii) and (iv) in (i)

$\Delta G = 820 -1350 = -530 J$
Hence, the change in Gibbs energy is

$-530$ J.

Calculate the free energy change at

$298 K$ for the reaction :

$Br_2(l) + Cl_2(g) \rightarrow 2BrCl(g)$ . For the reaction

$\Delta H^o = 29.3 kJ$ & the entropies of

$Br_2(l), Cl_2(g)$ & BrCl(g) at the

$298 K$ are

$152.3, 223.0, 239.7$

$J mol^{1} K^{1}$ respectively

0%
$1721.8 J$
0%
$1721.8 kJ$
0%
$17218 J$
0%
None of these

Explanation

$Br_2(l) + Cl_2(g) \rightarrow 2BrCl(g) , \Delta H^o = 29.3 kJ$

$S_{Br} = 152.3 S_{Cl_2(g)} = 223.0$

$S_{BrCl(g)} = 239.7 J mol^{1} K^{1}$

The entropy change for the reaction is

$\Delta S_R =2S_{BrCl(g)}- [S_{Br}+S_{Cl_2(g)}]=2 \times 239.7-[ 223.0+ 152.3] = 104.4 \: J mol^{1} K^{1}$

The relationship between the free energy change, the enthalpy change and the entropy change is as shown.

$\Delta_rG = \Delta H- T\Delta S$

Substitute values in the above expression.

$\Delta_rG = 29300 298 \times 104.4 = 1721.8 J$

Hence, the free energy change for the reaction is

$1721.8 J$

What is internal energy change of combustion for methanol?

0%
$-680\:kJ/mole$
0%
$-6.8\:kJ/mole$
0%
$-567\:kJ/mole$
0%
$-5.67\:kJ/mole$

Explanation

When 1 g of ice melts, the amount of heat absorbed is 340 J. This is as per

$\displaystyle \Delta H _{fusion}$ .

The volume contraction associated with melting of 1 g of ice is

$\displaystyle 1.091-1=0.091$ ml.

Thus 340 J heat corresponds to 0.091 ml of volume contraction.

The volume contraction of 1.82 ml correspond to

$\displaystyle \frac {340}{0.091} \times 1.82 = 6800$ J or

$6.8 kJ$ .

Since combustion is an exothermic reaction, there is decrease in the internal energy. Due to this, the change in internal energy has negative sign. Hence, the internal energy change of combustion for methanol is

$- 6.8 kJ/mol$ .

For the reaction

$X_2O_4(l)\rightarrow 2XO_2(g)$ ,

$\Delta U = 2.1\,Kcal,$ and

$\Delta S = 20\,cal\,K^{ -1 }$ at 300 K. Hence,

$\Delta G$ is:

0%
$2.7\,Kcal$
0%
$-2.7\,Kcal$
0%
$9.3\,Kcal$
0%
$-9.3\,Kcal$

Explanation

As we know,

$\Delta H = \Delta U + \Delta n_gRT$

$= 2.1\times 2\times 0.002\times 300 = 3.3\,Kcal$

$\Delta G = \Delta H - T\Delta S$

$= 3.3\times -300\times (0.02) = -2.7\,Kcal$

Hence, the correct option is

$\text{B}$

The energy of a system available to do work is called as:

0%
gibbs free energy
0%
heat of formation
0%
specific heat
0%
heinsenberg uncertainty principle
0%
heat of vaporization

What is the amount of the heat necessary to raise the temperature of

$50.0$ grams of liquid water from

$10.0 ^{\circ} C$ to

$30.0 ^{\circ}C$ .

(The specific heat of water liquid is

$4.18\ J/g/ ^{\circ}C$ )

0%
$20\ J$
0%
$80\ J$
0%
$100\ J$
0%
$200\ J$
0%
$4,180\ J$

Explanation

The amount of the heat necessary to raise the temperature of 50.0 grams of liquid water from 10.0 deg C to 30.0 deg

$^oC$ is obtained by multiplying the mass of water with a specific heat of water and the temperature change.
It is

$\displaystyle Q = mS\Delta T = 50.0 \times 4.18 \times (30.0-10.0) = 4180 J$
Note : The specific heat of water liquid is 4.18 J/g /

$^o$ C.

A certain amount of potassium chlorate on thermal decomposition gives oxygen which is sufficient for the combustion of ethane. When the products are cooled, the volume of the gaseous product is v ml. Identify the correct sequence of steps for the calculation of the mass of potassium chlorate
(a) Calculation of oxygen required for the combustion of ethane.
(b) Calculation of the amount of the ethane subjected to combustion from the volume of gaseous product.
(c) Calculation of potassium chlorate which can give the required amount of oxygen.
(d) Identification of the products of combustion of ethane and the product left after the cooling of products.

0%
a b c d
0%
d b a c
0%
b c a d
0%
d a c b

Explanation

Thermal decomposition of

$KCl{O}_{3}$

$2KCl{ O }_{ 3 }\left( s \right) \rightarrow 2KCl\left( s \right) +3{ O }_{ 2 }\left( g \right)$

Combustion of ethane

${ C }_{ 2 }{ H }_{ 6 }\left( g \right) +\cfrac { 7 }{ 2 } { O }_{ 2 }\left( g \right) \rightarrow 2C{ O }_{ 2 }\left( g \right) +3{ H }_{ 2 }O$

To find the mass of potassium chlorate

First we will identify the products of combustion of ethane and the products left after cooling of products

We required the products left after cooling so that the mass of the product is measured correct

From the volume of gases formed in the combustion of ethane, we can find the mass of ethane

From the mass of ethane we can find the amount of oxygen which is required to form products

Since the oxygen required for combustion of ethane is formed from the thermal decomposition of

$KCl{O}_{3}$ . So the amount of oxygen used in combustion of ethane is the same which is formed from the thermal decomposition of

$KCl{O}_{3}$ .

From the amount of oxygen, we can find the mass of potassium chlorate.

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 10 - MCQExams.com

Born-Haber Cycle

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 10 - MCQExams.com

Born-Haber Cycle

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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