Assertion (A): At equilibrium $$\Delta G$$ becomes zero. Reason (R): At equilibrium the two tendencies $$\Delta H$$ and $$T\Delta S$$ become equal.

Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Assertion is true but Reason is false.

Assertion is false and Reason is true.

MCQ Questions for Class 11 Medical Chemistry Thermodynamics Quiz 9

Which of the following parameters cannot be estimated by using the Born-Haber cycle?

0%
Hydration energy of the ion
0%
Electron gain enthalpy
0%
Lattice energy
0%
Electronegativity

Born-Haber cycle can be used to estimate :

0%
Lattice energy of ionic crystals
0%
Electron gain enthalpy
0%
Electronegativity
0%
Both (A) and (B)

Which one of the following gases possesses the largest internal energy?

0%
2 moles of helium occupying $1m^{3}$ at 300 K
0%
56 kg of nitrogen at 107 $Nm^{-2}$ and 300 K
0%
8 grams of oxygen at 8 atm and 300 K
0%
$6\times 10^{26}$ molecules of argon occupying $4m^{3}$ at 900 K

Which of the following relationship is not correct for the relation between

$\Delta H$ and

$\Delta U$ ?

0%
When $\Delta n_{g} = 0$ then $\Delta H = \Delta U$
0%
When $\Delta n_{g} > 0$ then $\Delta H > \Delta U$
0%
When $\Delta n_{g} < 0$ then $\Delta H < \Delta U$
0%
When $\Delta n_{g} = 0$ then $\Delta H > \Delta U$

Explanation

We know,

$\Delta H=\Delta U+\Delta n_{g}RT$

When

$\Delta n_{g} = 0$ then

$\Delta H=\Delta U$

For the electrochemical cell,

$\mathrm{M} | \mathrm{M}^{+} \| \mathrm{X}^{-} \mid \mathrm{X},$

$E_{M}^{o}+/ M=0 \cdot 44 V$ and

$E_{x / x^-}^{\circ}=0 \cdot 33 V$ From these data one can deduce that

0%
$M+X \rightarrow M^{+}+X^{-}$ is the spontaneous reaction
0%
$M^{+}+X^{-} \rightarrow M+X$ is the spontaneous reaction
0%
$E_{c e l l}=0 \cdot 77 V$
0%
$\mathrm{E}_{\mathrm{cell}}=-0.77 \mathrm{V}$

Explanation

$\mathrm{M}^{+}+e^{-} \rightarrow \mathrm{M} ; \quad \mathrm{E}^{\circ}=+0 \cdot 44 \mathrm{V} \dots(i)$

$X+e^{-} \rightarrow X^{-} ; \quad E^{\circ}=+0.33 \mathrm{V} \dots(ii)$
Subtracing Equ. (ii) from Eqn. (i), we have

$M^{+}-X \rightarrow M-X^{-} ; \quad E^{\circ}=+0 \cdot 11 V$
or

$\quad M^{+}+X^{-} \rightarrow M+X ; \quad E^{\circ}=+0 \cdot 11 V$
i.e.,

$\quad M^{+}+X^{-} \rightarrow M+X$ is the spontaneous reaction.

Hess's law is applicable for the determination of heat of

0%
transition
0%
formation
0%
reaction
0%
all of these

Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below

$\displaystyle \frac{1}{2}\mathrm{C}1_{2}(\mathrm{g})$ to

$\mathrm{C}1^{-}(\mathrm{a}\mathrm{q})$
(using the data,

$\Delta_{\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{s}}\mathrm{H}_{\mathrm{C}1_{2}}^{\Theta}=240\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}1^{-1},\ \Delta_{\mathrm{e}\mathrm{g}}\mathrm{H}_{\mathrm{C}1}^{\Theta}=-349\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}1^{-1},\ \Delta_{\mathrm{h}\mathrm{y}\mathrm{d}}\mathrm{H}_{\mathrm{C}1^{-}}=-381\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}1^{-1}$ ) will be

0%
$-850\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}1^{-1}$
0%
$+120\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}1^{-1}$
0%
$+152\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}1^{-1}$
0%
$-610\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}1^{-1}$

Explanation

Oxidising power of chlorine in aqueous solution can be estimated by calculating the overall enthalpy change of reaction.

The reaction takes place with following steps:

$1/2Cl_{2} -> Cl$ ΔH1

$Cl -> Cl^{-}$ ΔH2

$Cl^{-} -> Cl^{-} (aq)$ ΔH3

ΔH will be the sum of the energy involved in these

steps.

Hence, ΔH = ΔH1 +ΔH2 +ΔH3

ΔH1 = 240/2 = 120kJ/mol, ΔH2 = -349kJ/mol, ΔH3 = -381kJ/mol

Hence, ΔH = 120-349-381 = -610kJ/mol

Standard enthalpy and standard entropy for the oxidation of $NH_3$ at $298$ K are $-382.64$ $\text{kJ}\ \text{mol}^{-1}$ and $-145.6$ $\text{J}\ {\text{mol}}^{-1}\ \text K^{-1}$ respectively. Standard free energy change (in $\text{kJ mol}^{-1}$ ) for the same reaction at $298$ K is:

0%
$-221.1$
0%
$-339.3$
0%
$-439.3$
0%
$-523.2$

Explanation

We know the relation:

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

$\Delta H=-382.64$ kJ mol

$^{-1}$ (given)

$\Delta S=-145.6$ J mol

$^{-1}$ K

$^{-1}$ (given)

$\Delta G=-382.64+298\times 145.6\times 10^{-3}$ kJ mol

$^{-1}$

$=-339.25$ kJ mol

$^{-1}$

$\Delta G\simeq -339.3$ kJ mol

$^{-1}$

Hence, the free energy change for the reaction at

$298$ K is

$-339.3$ kJ mol

$^{-1}$ .

Option B is correct.

For the equilibrium :

$CaC{O}_{3}\left(s\right) \rightleftharpoons CaO\left(s\right) + C{O}_{2}\left(g\right)$ , the pressure of

$C{O}_{2}$ is 1 atm at equilibrium in a container of

$V$ litre, at temperature

$T$ . Which one is not correct?

0%
$T = \displaystyle\frac{\Delta H}{\Delta S}$
0%
$\Delta G = 1$
0%
$\Delta G = 0$
0%
$T = \displaystyle\frac{V}{nR}$

Explanation

At equilibrium, there is no change in free energy.

$\Delta G=0$ at equilibrium.

Hence, option B is correct

Assertion (A): At equilibrium $\Delta G$ becomes zero.

$T\Delta S$ become equal.

0%
Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
0%
Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
0%
Assertion is true but Reason is false.
0%
Assertion is false and Reason is true.

Explanation

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

at

$eq^{m}\Delta G=O$

$\therefore \Delta H=T\Delta S$

Also at

$eq^{m}\Delta H=T\Delta S$

$\therefore \Delta G=O$

The standard Gibbs energy change for the formation of propane $(C_3H_{8(g)})$ at 298 K is

$[\Delta _{f}H^{\theta }$ for propane $=-103.85$ kJ mole $^{-1},$ $S^{\theta }_{m}\;C_{3}H_{8(g)}=270.2$ JK $^{-1}$ mol $^{-1}$ , $S^{\theta }_{m}\;H_{2(g)}=130.68$ JK $^{-1}$ mol $^{-1}$ , $S^{\theta}_{m}C_{(graphite)}=5.742$ JK $^{-1}$ mol $^{-1}]$

0%
$-23.4$ kJ
0%
$-44.4$ kJ
0%
$-54.4$ kJ
0%
$-104.5$ kJ

Explanation

Formation of propane

$3C + 4H_2 \rightarrow C_3H_8$

$\Delta{S}_f$ for propane

$= 270.2 - 4\times 130.68 - 3\times 5.742 = -269.746$ JK

$^{-1}$ mol

$^{-1}$

$\Delta{H} = -103.85$ kJK

$^{-1}$ mol

$^{-1}$

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

$\Delta{G} = -103.85 - 298\times(-269.746)/1000 = -23.4$ kJ

Match the list -I with list - II.

List-I List - II

A) Hess law is not applicable for 1) is not a state function

B) All combustion reactions are 2) Heat of dilution

C) Work 3) Exothermic

D) Difference between two integral 4) Nuclear reaction

heats of solutions

0%
A-1, B-2, C-3, D-4
0%
A-4, B-3, C-1, D-2
0%
A-3, B-2, C-4, D-1
0%
A-2, B-1, C-4, D-3

Standard entropy of $X_{2}$ , $Y_{2}$ and $XY_{3}$ are 60, 40 and 50 $JK^{-1}mol^{-1}$ respectively. For the reaction: $\dfrac{1}{2}X_{2}+\dfrac{3}{2}Y_{2}\rightarrow XY_{3}, \Delta H=-30\: kJ$ to be at equilibrium, the temperature will be :

0%
1250 K
0%
500 K
0%
1000 K
0%
750 K

Explanation

At equilibrium,

$\Delta G=0=\Delta H-T\Delta S$

$\implies \Delta H=T\Delta S$

Now,

$\Delta S=50-\dfrac{60}{2}-\dfrac{40}{2}\times 3$

$\Delta S=-40\: JK^{-1}$

Again,

$\implies -30\times 1000=T\times (-40)$

$\therefore T=750K$

The value $\Delta$ S for the process, $H_{2}O(s) \rightarrow H_{2}O(l)$ at 1 atm pressure and 260K is greater than zero. The value of $\Delta G$ will be :

0%
$>0$
0%
$<0$
0%
$=0$
0%
lies between $-1$ and $0$

Explanation

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

$\Delta S$ + ve and

$\Delta H$ + ve.

The temperature is low so the process is non-spontaneous.

$\therefore \Delta G> 0$ .

A coffee cup calorimeter initially contains 125g of water at a temperature of $24.2^{0}C$ . After 10.5gm solute is added to the water at the same temperature, the final temperature becomes $21.1^{0}C$ . The heat of solution is:

0%
85 J/g
0%
110 J/g
0%
270 J/g
0%
167 J/g

Explanation

Heat released

$= (125+10.5) \times (24.2 - 21.1)$

$= 420.05$ cal

Heat of solution

$=\dfrac{420.05}{10.5} {cal}\ {g}^{-1}$ [Note: 1 cal

$=$ 4.186 J]

$=40\ cal/g$

Heat of solution

$=$ 167 J/g

The heat of combustion for $C$ , $H_{2}$ and $CH_{4}$ are $-349$ kJ/mol, $-241.8$ kJ/mol and $-906.7$ kJ/mol respectively. The heat of formation in kJ/mol of $CH_{4}$ is :

0%
$174.1$
0%
$274.1$
0%
$374.1$
0%
$74.1$

Explanation

Given that,

$C+O_{2}\rightarrow CO_{2}(g)\ ; \ \Delta H=-349$ kJ/mol ..........

$(i)$

$H_{2}+\frac{1}{2}O_{2}\rightarrow H_{2}O(l)\ ;\ \Delta H=-241.8$ kJ/mol ..........

$(ii)$

$CH_{4}+ 2O_{2}\rightarrow CO{_{2}}{(g)}+2H_{2}O(l)\ ; \ \Delta H=-906.7$ kJ/mol .........

$(iii)$

Now, we perform

$(i) + 2 \times (ii) - (iii)$ , we get

$C(s)+2H_{2}(g)\rightarrow CH_{4}(g)$

Therefore,

$\Delta H=-349+2(-241.8)+906.7$

$=74.1$ kJ/mol

Hence, the heat of formation of

$CH_4$ is

$74.1$ kJ/mol.

Select the incorrect statement(s) :

0%
Lattice energy $\propto \frac{1}{r^2}$ {where r is the interionic distance}.
0%
Lattice energy $\propto$ $q_1q_2$ . {where $q_1$ and $q_2$ are the charges of co-ions}.
0%
Ionic mobility in aqueous state $\propto \frac{1}{radius\ of\ ions\ in\ gaseous\ state}$
0%
Heat of formation of a compound depends on the number of steps involved in its formation reaction.

The change in free energy accompanied by the isothermal reversible expansion of 1 mole of an ideal gas when it doubles its volume is

$\Delta G_{1}$ . The change in free energy accompanied by a sudden isothermal irreversible doubling volume of 1 mole of the same gas is

$\Delta G_{2}$ . The ratio of

$\Delta G_{1}$ and

$\Delta G_{2}$ is :

0%
$1$
0%
$\dfrac{1}{2}$
0%
$2$
0%
$\dfrac{3}{2}$

Explanation

The Gibb's free energy of the system is a state function.

Moving from state

$1$ to

$2, \: \Delta G$ will be the same irrespective of the type of process.

$\therefore \dfrac{\Delta G_{1}}{\Delta G_{2}}=1$

Consider the following processes.

	$\Delta H (kJ/mol)$
$1/2A\rightarrow B$	+150
$3B\rightarrow 2C+D$	-125
$E+A\rightarrow 2D$	+350

For

$B+D\rightarrow E+2C, \Delta H$ will be:

0%
325 kJ/mol
0%
525 kJ/mol
0%
-175 kJ.mol
0%
-325 kJ/mol

Explanation

$2(i)-(iii)+(ii)$

$\Delta H=2(150)-350-125$

$=-175kJ/mol$

In conversion of lime-stone to lime,

$CaCO_{3}(s)\rightarrow CaO(s) + CO_{2}(g)$ , the values of

$\Delta^0H$ and

$\Delta S^0$ are

$+179.1\ kJ mol^{-1}$ and

$160.2\ J/Kmol$ respectively at

$298\ K$ and

$1$ bar. Assuming that

$\Delta H^o$ and

$\Delta S^o$ do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is:

0%
$1008\ K$
0%
$1200\ K$
0%
$845\ K$
0%
$1118\ K$

Explanation

$\textbf{Explanation for correct option:}$

$\textbf{Step 1: Criteria of spontaneity }$

Gibbs free energy is used to check the spontaneity of a chemical reaction.

$\Delta{G}=-ve$ , the reaction is spontaneous.

$\Delta{G}=+ve$ , the reaction is non-spontaneous.

$\textbf{Step 2: Calculation for temperature }$

$\Delta{G}=\Delta{H}-T\Delta{S}$ .............(1)

Put the value of

$\Delta{H}$ and

$\Delta{S}$ in equation (1)

For a spontaneous reaction

$\Delta{G}=-ve$

$0> 179.1kJmol^{-1}-T\times {160.2JK^{-1}mol^{-1}}$

$T> \dfrac{179.1\times 10^3Jmol^{-1}}{{160.2Jmol^{-1}}}$

$T> 1118K$

The values of

$\Delta H$ and

$\Delta S$ for the reaction, C(graphite) +

$CO_2(g)\rightarrow 2CO(g)$ are 170 kJ and

$170\; JK^{-1}$ respectively. This reaction will be spontaneous at:

0%
510 K
0%
710 K
0%
910 K
0%
1010 K

Explanation

For a spontaneous reaction,

$\triangle G=$ Negative

And at equilibrium

$\triangle G=0$ i.e

$\triangle H=T\triangle S$

$\Rightarrow T=\cfrac {170\times 10^{3}}{170}$

$\Rightarrow T=1000K$

Therefore, at this temperature, reaction is at equilibrium i.e

$\triangle G=0$ and above this temperature

$\triangle G$ will be negative and reaction will be spontaneous.

A reaction occurs spontaneously if:

0%
T $\Delta$ S > $\Delta$ H and $\Delta$ H is +ve and $\Delta$ S is -ve
0%
T $\Delta$ S = $\Delta$ H and both $\Delta$ H and $\Delta$ S are +ve
0%
T $\Delta$ S < $\Delta$ H and both $\Delta$ H and $\Delta$ S are +ve
0%
T $\Delta$ S > $\Delta$ H and both $\Delta$ H and $\Delta$ S are +ve

Explanation

Hint: $\Delta G <0$ is a necessary and sufficient condition.

Step 1: Write the formula for $\Delta G$

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

Step 2: Equate $\Delta G<0$ as the condition for spontaneity.

$\Delta H-T\Delta S < 0$

$\implies T\Delta S > \Delta H$

But if

$\Delta H \rightarrow +ve, \Delta S \rightarrow -ve,$

$T\Delta S>\Delta H$ Is not possible since the right hand side is positive which cannot be less than the negative left hand side. So we discard option

$A$ .

Final answer: Option $D$

A 1000 gm sample of water is reacted with an equimolar amount of CaO (both at same initial temperature of

$25^{\circ}C$ ). Assuming the container to be adiabatic, the final temperature of the product is approx :

Given :

$CaO + H_2O\rightarrow Ca(OH)_2,\ \Delta H=-65.2 KJ/mol$

Specific heat of

$Ca(OH)_2 = 1.2 \: J/gm^{\circ}C$

0%
$735^{\circ}C$
0%
$760^{\circ}C$
0%
$746^{\circ}C$
0%
$789^{\circ}C$

Explanation

$\text{Total heat released}=65.2\times 10^3\times \frac{1000}{18}$

$\text{Mass of }Ca(OH)_2 \text{produced} = \frac{1000}{18}\times74 \: gm$

Now, by energy balance equation:

$\frac{1000}{18}\times74 \times 1.2(T_f-25)=\frac{1000}{18}\times 65.2\times 10^3$

$T_f=759.23^0C\approx 760^0C$

$\Delta H{_{f}}^{0}$ for

$Ag^{+} (\infty$ diluted),

$NO{_{3}}^{-} (\infty$ diluted),

$Cl^{-} (\infty$ diluted) and

$AgCl_{(S)}$ are

$-105.579, -207.36, -167.159$ and

$-127.068$ respectively. Calculate the enthalpy change for the reaction

$AgNO_{3(aq)}+HCl_{(aq)}\rightarrow AgCl_{(S)}+HNO_{3} (aq)$ .

0%
21.471 kJ/mol
0%
145.67 kJ/mol
0%
65.488 kJ/mol
0%
None

Explanation

$\Delta H^{\circ}_{reaction}=[\Delta H_{f}(AgCl)+\Delta H_{f}(H^{+})+\Delta H_{f}(NO{_{3}}^{-})]$

$- [ \Delta H^{\circ}_{f}(Ag^{+}) +\Delta H^{\circ}_{f}(NO{_{3}}^{-})+\Delta H^{\circ}_{f}(Cl^{-})+\Delta H^{\circ}_{f}(H^{+})]$

$\Delta H^{\circ}_{reaction}=-127.068-\left [- 105.579-167.159 \right ]$

$\Delta H^{\circ}_{reaction}=145.67$ kJ/mol.

$C(s) + O_2(g) \rightarrow CO_2(g), \quad \Delta{H} = + 94.0 K cal.$

$CO(g) +\frac{1}{2} O_2(g) \rightarrow CO_2(g) ,\quad \Delta H = -67.7 K cal.$

From the above reactions find how much heat (Kcal mole

$^{-1}$ ) would be produced in the following reaction:

$C(s) + \frac{1}{2} O_2(g) \rightarrow CO(g)$

0%
20.6
0%
26.3
0%
44.2
0%
161.7

Explanation

The thermochemical reactions are as given below.

$C(s) + O_2(g) \rightarrow CO_2(g) \Delta H = + 94.0 K cal.$ ......(1)

$CO(g) +\frac{1}{2} O_2(g) \rightarrow CO_2(g), \Delta H = -67.7 K cal$ ......(2)

The reaction (2) is reversed and added to the reaction (1) to obtain the reaction

$C(s) + \frac{1}{2} O_2(g) \rightarrow CO(g)$ .

The heat produced in this reaction is

$94.0-(-67.7)=161.7$ Kcal/mole

Given that

$Zn + \frac{1}{2} O_2 \rightarrow ZnO\quad\Delta{H}= + 84000 cal$ ........... 1

$Hg+\frac{1}{2} O_2 \rightarrow HgO\quad \Delta{H} =+ 21700 cal$ ........... 2

The heat of reaction

$(\Delta H)$ for,

$Zn + HgO \rightarrow ZnO + Hg$ is

0%
105700 cal
0%
62300 cal
0%
-105700 cal
0%
-62300 cal

Explanation

The thermochemical reactions are as given below.

$Zn + \frac{1}{2} O_2 \rightarrow ZnO + 84000 cal$ ......1

$Hg+\frac{1}{2} O_2 \rightarrow HgO + 21700 cal$ ......2

The reaction 2 is reversed and added to the reaction 1 to obtain the reaction

$Zn + HgO \rightarrow ZnO + Hg$

The heat of this reaction is

$84000-21700=62300$

Specific heat of substance at m.pt. (or b.pt. or for isothermal process) and specific heat of a substance during adiabatic change respectively are

0%
0, 0
0%
0, $\infty$
0%
$\infty$ , 0
0%
$\infty, \infty$

Explanation

The amount of heat required to raise the temperature of unit mass of a substance by one degree celsius is known as specific heat. It is given by the ratio

$\displaystyle \left ( \frac{\partial H}{\partial T} \right )$
When at constant pressure, one phase of a substance is in equilibrium with other phase (during melting or boiling), the heat

$\partial H$ is supplied to the system, and the temperature change

$\partial T$ is zero as the heat supplied is utilized for the phase transition.
Hence, the specific heat is infinite. This is also true for the isothermal process.
During adiabatic process, no heat is supplied to system. The term

$\partial H$ is zero. Hence the specific heat is zero.

The heat of reaction for

$A + \frac{1}{2} O_2 \rightarrow AO$ is

$-50$ kcal/mol and

$AO + \frac{1}{2} O_2 \rightarrow AO_2$ is

$100$ kcal/mol.
The heat of reaction (in kcal/mol) for

$A + O_2 \rightarrow AO_2$ will be:

0%
$-50$
0%
$+50$
0%
$+100$
0%
$+150$

Explanation

The thermochemical reactions are given below:

$A + \dfrac{1}{2} O_2 \rightarrow AO \ ; \Delta H =-50$ kcal/mol

$AO + \dfrac{1}{2} O_2 \rightarrow AO_2\ ; \Delta H =100$ kcal/mol

The above two reactions are added to obtain the reaction as follows:

$A + O_2 \rightarrow AO_2$

Therefore, the heat of the reaction is

$-50+100=50$ kcal/mol.

For an ideal gas, enthalpy becomes:

0%
$\Delta H = \Delta U - RT$
0%
$\Delta H = -\Delta U - RT$
0%
$\Delta H = \Delta U + RT$
0%
$\Delta H = -\Delta U + RT$

Consider the following data:

$\Delta_fH^o(N_2H_4, l)=50kJ/mol, \Delta_fH^o(NH_3, g)=-46 kJ/mol$ ,

$B.E. (N-H)=393 kJ/mol$ and

$B.E. (H-H)=436 kJ/mol$ ,

$\Delta_{vap}H(N_2H_4, l)=18 kJ/mol$
Calculate the

$N-N$ bond energy in kJ/mol for

$N_2H_4$ .

0%
$190 \ kJ/mol$
0%
$-190 \ kJ/mol$
0%
$95 \ kJ/mol$
0%
$-95 \ kJ/mol$

Following are the thermochemical reactions:

$C$ (graphite)

$+ \dfrac{1}{2}O_2 \rightarrow CO \ ; \Delta H = - 110.5$ kJ/mol

$CO + \dfrac{1}{2} O_2 \rightarrow CO_2 \ ; \Delta H = - 283.2$ kJ/mol

The heat of reaction (in kJ/mol) for the following reaction is:

$C (\text{graphite}) + O_2 \rightarrow CO_2$

0%
$+393.7$
0%
$-393.7$
0%
$-172.7$
0%
$+172.7$

Explanation

The thermochemical reactions are as given below:

$C$ (graphite)

$+ \dfrac{1}{2}O_2 \rightarrow CO ; \Delta H = - 110.5$ kJ/mol

$...(1)$

$CO + \dfrac{1}{2} O_2 \rightarrow CO_2 ; \Delta H = - 283.2$ kJ/mol

$......(2)$
The above two reactions are added to obtain the following reaction:

$C$ (graphite)

$+ O_2 \rightarrow CO_2$
By applying Hess's law, the enthalpy change will be

$- 110.5+(- 283.2)=-393.7$ kJ/mol

Hence, the heat of the given reaction is

$-393.7$ kJ/mol.

Find the heat change in the reaction

$NH_3(g) + HCl (g) \rightarrow NH_4 Cl(s)$ from the following data

$NH_3 (g) + aq \rightarrow NH_3 (aq),$

$\Delta H = - 8.4 K. Cal.$

$HCl(g) + aq \rightarrow HCl(aq),$

$\Delta H = -17.3 K. Cal.$

$NH_3(aq) + HCl(aq) \rightarrow NH_4 Cl(aq), \Delta H = - 12.5 K. Cals.$

$NH_4 Cl(s) + aq \rightarrow NH_4 Cl(aq), \Delta H = + 3.9 K. Cal.$

0%
-42.1
0%
-23.3
0%
+34.3
0%
+42.1

Explanation

The thermochemical reactions are as given below.

$NH_3 (g) + aq \rightarrow NH_3 (aq),\Delta H = - 8.4 K. Cal.$

$HCl(g) + aq \rightarrow HCl(aq),\Delta H = -17.3 K. Cal.$

$NH_3(aq) + HCl(aq) \rightarrow NH_4 Cl(aq), \Delta H = - 12.5 K. Cal.$

$NH_4 Cl(s) + aq \rightarrow NH_4 Cl(aq), \Delta H = + 3.9 K. Cal.$
The reaction (4) is reversed and added to the reactions (1), (2) and (3) to obtain the reaction

$NH_3(g) + HCl (g) \rightarrow NH_4 Cl(s)$ .
Hence, the heat change of the reaction is

$-8.4-17.3-12.5-3.9=-42.1$ K. Cal

For the reaction :

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

$\Delta U = 2.1\,kcal,\ \Delta S = 20\,cal\,K^{ -1 }\,at\ 300\,K$ .
Hence

$\Delta G$ is:

0%
$-7\,kcal$
0%
$-4.7\,kcal$
0%
$+2.7\,kcal$
0%
$-2.7\,kcal$

Explanation

As we know,

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

$= 2.1+ 2\times 0.002\times 300 = 3.3\,kcal$

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

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

200 ml of

HCl is mixed with

400 ml of

0.5M

NaOH. The temperature rise in the calorimeter was found to be

4.0oC. Water equivalent of calorimeter is

25g and the specific heat of the solution is

1 cal/mL/degree. If the theoritical heat of neutralization of a strong acid and strong base is

−13.5 kcal, then the percentage error in this experiment while calculating the heat of neutralization is

0%
7.4
0%
3
0%
5.6
0%
4.5

Explanation

Total volume

$200+400=600 ml$
200 ml of 1 M acid

$=$ 400 ml of 0.5 M

$NaOH$
200 m. eq. of acid reacts with 200 m.eq. of base

$=\Delta H$
Heat of neutralization

$=5\times \Delta H$
Heat produced during neutralization

$=$ heat taken up by calorimeter + solution

$=m_{1}s_{1}\Delta T+m_{2}s_{2}\Delta T$

$=(25\times 1 \times 4)+(600\times 1 \times 4)=2500 cal$

$\therefore$ Heat of neutralization

$=-5\times 2500$

$=-12500 cal=-12.5 kcal$
% error

$=(\dfrac{13.5-12.5}{13.5})\times 100=7.4$

For the hypothetical reaction:

${ A }_{ 2 }(g)+{ B }_{ 2 }(g)\rightleftharpoons 2AB(g)$

${ \Delta }_{ r }{ G }^{ o }$ and

${ \Delta }_{ r }{ S }^{ o }$ is

$20 kJ/mol$ and

$-20{ JK }^{ -1 }{ mol }^{ -1 }$ respectively at

$200K$ . If

${ \Delta }_{ r }{ C }_{ p }$ is

$20{ JK }^{ -1 }{ mol }^{ -1 }$ then

${ \Delta }_{ r }{ H }^{ o }$ at

$400K$ is :

0%
$20\ kJ/mol$
0%
$7.98\ kJ/mol$
0%
$28\ kJ/mol$
0%
none of these

Explanation

As we know,

$\Delta G =\Delta H- T \Delta S; \Rightarrow 20=\Delta H -200\times \dfrac{-20}{1000}; \Rightarrow \Delta H = 16kJ/mol$

Also,

$\Delta H= \Delta_r C_p (T_2 - T_1); \Delta H = 16+20\times \dfrac{200}{1000} = 20kJ/mol$

Using the listed information calculate

${ \Delta }_{ r }{ G }^{ o }$ (in kJ/mol) at

${ 27 }^{ o }C$ .

${ Co }_{ 3 }{ O }_{ 4 }(s)+4CO(g)\longrightarrow 3Co(s)+4{ CO }_{ 2 }(g)$

Given : At

$300K$ ,

$\Delta { { H }^{ o } }_{ f }(kJ/mol)$ are

$-891,\ -110.5,\ 0.0$ and

$-393.5$ respectively.

${ S }^{ o }(J/K.mol)$ are

$102.5,\ 197.7, \ 30.0$ and

$213.7$ respectively.

0%
$214.8$
0%
$-195.0$
0%
$-200.3$
0%
$-256.45$

0.16 g of methane was subjected to combustion at

$27^{\circ}C$ in bomb calorimeter. The temperature of calorimeter system (including water) was found to rise by

$0.5^{\circ}C$ . If the heat of combustion of methane at constant volume and constant pressure is

$x$ kJ/mole, find the value of

$x$ . The thermal capacity of the calorimeter system is

$17.7\ kJ\ K^{-1},\ R=8.314\ JK^{-1}\ mol^{-1}$ .

0%
$-890$
0%
$890$
0%
$450$
0%
$-450$

Explanation

Heat given by burning of 0.16 g methane = heat taken up by calorimeter =

$17.7 \times 0.5 =8.85 \:kJ$

$\displaystyle \therefore$ Heat given by burning of 16 g methane

$=\dfrac{8.85\times 16}{0.16}=885 \:kJ$

The burning is taking place at constant volume in a bomb calorimeter.

$\therefore \Delta U= -885 \: kJ \: mol^{-1}$

$\because \Delta H= \Delta U+\Delta nRT$

$CH_4+ 2O_2\longrightarrow CO_2 + 2H_2O(l)$ ;

$\therefore \Delta n = -2$

$\therefore \Delta H= -885+(-2)\times 8.314\times 10 ^{-3} \times 300$

$= -889.98\: kJ \: mol^{-1}$

Hence, the correct option is

$\text{A}$

Determine

${ \Delta }{ U }^{ o }$ at

$300K$ for the following reaction using the listed enthalpies of reaction:

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

$C_{(graphite)}+1/2{ O }_{ 2 }(g)\longrightarrow CO(g);\quad \Delta { { H }_{ 1 } }^{ o }=-110.5kJ$

$CO(g)+1/2{ O }_{ 2 }(g)\longrightarrow { CO }_{ 2 }(g);\quad \Delta { { H }_{ 2 } }^{ o }=-282.9kJ$

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

$C_{(graphite)}+2{ H }_{ 2 }(g)\longrightarrow { CH }_{ 4 }(g);\quad \Delta { { H }_{ 4 } }^{ o }=-74.8kJ$

0%
$653.5\ kJ$
0%
$-686.2\ kJ$
0%
$-747.4\ kJ$
0%
None of these

The enthalpy changes of the following reactions at

$27^{\circ}C$ are

$\displaystyle \:Na(s)+\frac{1}{2}Cl_{2}(g)\rightarrow NaCl(s)$	$\displaystyle \:\Delta _{r}H= -411 kJ/mol$
$\displaystyle \:H_{2}(g)+S(s)+2O_{2}(g)\rightarrow H_{2}SO_{4}(l)$	$\displaystyle \:\Delta _{r}H= -811 kJ/mol$
$\displaystyle \:2Na(s)+S(s)+2O_{2}(g)\rightarrow Na_{2}SO_{4}(s)$	$\displaystyle \:\Delta _{r}H= -1382 kJ/mol$
$\displaystyle \:\frac{1}{2}H_{2}(g)+\frac{1}{2}Cl _{2}(g) \rightarrow HCl(g)$	$\displaystyle \:\Delta _{r}H= -92 kJ/mol;$

from these data, the heat change of reaction at constant volume (in kJ/mol) at

$27^{\circ}C$ for the process

$\displaystyle \:2NaCl(s)+H_{2}SO_{4}(l)\rightarrow Na_{2}SO_{4}(s)+2HCl(g)$

0%
67
0%
62.02
0%
71.98
0%
None

Explanation

$\Delta H_{reaction}=$

$2\Delta H_{f}HCl$ +

$\Delta H_{f}Na_{2}SO_{4}$

$-\Delta H_{f}Na_{2}SO_{4}$

$-2\Delta H_{f}NaCl$

$\Delta H_{reaction}=$ 2*(-92)+(-1382)-(--811)-(-2*411)

$\Delta H_{reaction}=$ = 67kJ/mol

Combustion of surose is used by aerobic organisms for providing energy for the life sustaining processes. If all the capturing of energy from the reaction is done through electrical process (non P-V work) then calculate maximum available energy which can be captured by combustion of

$34.2gm$ of sucrose
Given :

$\Delta {{H}_{combustion}}(sucrose)=-6000\ kJ. {mol}^{-1}$

$\Delta {{S}_{combustion}}=180J/K.mol$ and body temperature is

$300K$

0%
$600kJ$
0%
$594.6kJ$
0%
$5.4kJ$
0%
$605.4kJ$

Explanation

As we know,

$\Delta G = \Delta H - T \delta S =-6000-300\times 0.18 = -6054\ kJ/mol$

So, for 34.2gm (0.1 mol) of sucrose maximum available energy

$\Delta G = 605.4\ kJ$

Find the forward

$\displaystyle \:\Delta _{r}U^{\circ}$ at 300K for the reaction

$4HCl(g)+O_{2}(g)\rightarrow 2Cl_{2}(g)+2H_{2}O(g)$ Assume all gases are ideal. Given

$\displaystyle \:H_{2}(g)+Cl_{2}(g)\rightarrow 2HCl(g)$

$\Delta _{r}H_{300}^{^{\circ}}= -184.5 kJ/mol$

$\displaystyle \:2H_{2}(g)+O_{2}(g)\rightarrow 2H_2O(g)$

$\Delta _{r}H_{300}^{^{\circ}}= -483 kJ/mol$ (Use R= 8.3 J/mole)

0%
111.5 kJ/mole
0%
-109.01 kJ/mole
0%
-111.5 kJ/mole
0%
None

Explanation

$4HCl(g)+O_2(g) \longrightarrow 2Cl_2(g)+2H_2O(g)......(i)$

$H_2(g)+Cl_2(g) \longrightarrow 2HCl(g)....(ii)$

$2H_2(g)+O_2(g) \longrightarrow 2H_2O(g)....(iii)$
As it can be clearly seen

$Eq(i)=Eq(iii)-2Eq(ii)$
Thus

$\Delta H_r^{o} (i)=\Delta H_r^{o}(iii)-2\Delta H_r^{o}(ii)$

$=-483-(2 \times -184)$

$=-114kJ/mol$

$\Delta U_r^{o}=\Delta H_r^{o}-\Delta n_gRT$
Thus

$\Delta U_r^{o}=-114-(4-5) \times 0.0083 \times 300$

$\Rightarrow \Delta U_r^{o}=-111.5kJ/mol$

The reaction that proceeds in the forward direction is:

0%
$SnC{l}_{4} + H{g}_{2}C{l}_{2} \longrightarrow SnC{l}_{2} + 2HgC{l}_{2}$
0%
$N{H}_{4}Cl + NaOH \longrightarrow {H}_{2}O + N{H}_{3} + NaCl$
0%
${Mn}^{2+} + 2{H}_{2}O + C{l}_{2} \longrightarrow Mn{O}_{2} + 4{H}^{+} + 2C{l}^{-}$
0%
${S}_{4}{O}_{6}^{2-} + 2{I}^{-} \longrightarrow 2{S}_{2}{O}_{3}^{2-} + {I}_{2}$

In the Born-Haber cycle for the formation of solid common salt (

$Nacl$ ), the largest contribution comes from

0%
the low ionization potential of $Na$
0%
the high electron affinity of $Cl$
0%
the low $\Delta H_{vap}$ of $Na(s)$
0%
the lattice energy

Explanation

REF.Image

$\Delta _{Hf^{\circ }}=\Delta H_{sub}+IE+\Delta H_{diss}+EA+U$

$\Delta _{Hf^{\circ }}=108+496+122-349-788=-411 kJ/mole$

Use the following data to calculate second electron ainity of oxygen, i.e., for the process

$O^{-}(g) + e^{-}(g) \rightarrow O^{2-}(g)$
Is the

$O^{2-}$ ion stable in the gas phase?.Why is it stable in solid MgO?
Heat of sublimation of

$Mg(s) = + 147.7 kJ mol^{-1}$
Ionisation energy of Mg(g) to form

$Mg^{2+}(g) = + 2189.0 kJ mol^{-1}$
Bond dissociation energy for

$O_2 = + 498.4 kJmol^{-1}$
First electron affinity of

$O(g) = - 141.0 kJ mol^{-1}$
Heat formation of

$MgO(s) = -601.7 kJ mol^{-1}$
Lattice energy of

$MgO = -3791.0 kJ mol^{-1}$

0%
601.7
0%
744.4
0%
1346.1
0%
147.7

Explanation

Option (B) is correct.

$\triangle H_1=-1346.1+q\\ Mg(s)+\frac { 1 }{ 2 } O_{ 2 }(g)\rightarrow MgO(s)\\ \triangle H_{ 2 }=-601.7\\ By\quad Born-Haber\quad Cycle(based\quad on\quad Hess\quad law)\\ \triangle H_{ 1 }=\triangle H_{ 2 }\\ -1346.1+q=-601.7\\ \qquad \qquad q=744.4kJ\quad mol^{-1}$

The energy change for the alternating reaction that yields chlorine sodium $(Cl^{+}Na^{-})$ will be:

$2Na(s)\, +\, Cl_2(g)\,\rightarrow\, 2Cl^{+}Na^{-}(s)$

Given that:

Lattice energy of $NaCl\,=\,-787\, kJ\,mol^{-1}$

Electron affinity of $Na\,=\,-52.9\, kJ\, mol^{-1}$

Ionisation energy of $Cl\, =\, +\,1251\, kJ\, mol^{-1}$

BE of $Cl_2\,=\,244\, kJ\, mol^{-1}$

Heat of sublimation of $Na(s)\, =\,107.3\, kJ\, mol^{-1}$

$\Delta H_f(NaCl)\, =\,-411\, kJ\, mol^{-1}$ .

0%
+640 kJ
0%
+1280 kJ
0%
-410 kJ
0%
+410 kJ

Select correct statement.

0%
Both lattice energy and hydration energies decrease with ionic size.
0%
Lattice energy can be calculated using Born-Haber cycle.
0%
If the anion is larger compared to the cation, the lattice energy will remain almost constant within a particular group.
0%
All the above are correct statements.

The lattice energy of NaCl(s) using the following data will be:

heat of sublimation of $Na(s)\,=\,S$

$(IE)_1$ of $Na\,(g)\,=\,I$

bond dissociation energy of $Cl_2\,(g)\,=\,D$

electron affinity of $Cl\,(g)\,=\,-E$

heat of formation of

$NaCl(s)\,=\,-Q$

0%
Lattice energy $-U\, =\, S\, +\, I\, +\,\displaystyle \frac{D}{2}\, -\, E\, -\,Q$
0%
Lattice energy $-U\, =\, S\, -\, I\, +\,\displaystyle \frac{D}{2}\, -\, E\, -\,Q$
0%
Lattice energy $-U\, =\, S\, +\, I\, +\,\displaystyle \frac{D}{2}\, +\, E\, -\,Q$
0%
Lattice energy $-U\, =\, S\, -\, I\, -\,\displaystyle \frac{D}{2}\, +\, E\, +\,Q$

Explanation

An estimate of the strength of the bonds in an ionic compound can be obtained by measuring the lattice energy of the compound, which is the energy given off when oppositely charged ions in the gas phase come together to form a solid.

The Lattice energy of NaCl from its elements Sodium and Chlorine in their stable forms is modeled in five steps in the diagram:

Enthalpy change of atomization enthalpy of lithium
Ionization enthalpy of lithium
Atomization enthalpy of fluorine
Electron affinity of fluorine
Lattice enthalpy

$-U\, =\, S\, +\, I\, +\,\displaystyle \frac{D}{2}\, -\, E\, -\,Q$

An ideal monoatomic gas undergoes a process in which its internal energy U and density

$\rho$ vary as

$U\rho\, =\, constant.$ The ratio of change in internal energy and the work done by the gas is

0%
$\displaystyle \frac{3}{2}$
0%
$\displaystyle \frac{2}{3}$
0%
$\displaystyle \frac{1}{3}$
0%
$\displaystyle \frac{3}{5}$

Explanation

As we know that internal energy(U) of a monoatomic gas is given as

$\dfrac{{m}{v}^{2}}{2}$
and work done (W)=F.S
where

$F=\dfrac{{mn}{v}^{2}}{3Ln}$
and

$S=L$
so

$\dfrac{internal energy}{work done}$ =

$\dfrac{3}{2}$

The value of

$\Delta U$ for the reaction

$2A\left(g\right) + B\left(g\right) \rightleftharpoons {A}_{2}B\left(g\right)$ for which

${K}_{p} = 1.0\times {10}^{-10} {atm}^{-2}$ and

$\Delta S = 5 J{K}^{-1}$ and

$T = 300 K$ , is :

0%
53.93 kJ
0%
63.93 kJ
0%
56.24 kJ
0%
68.24 kJ

Explanation

As we know,

$\Delta G = -2.303 RT \log{{K}_{p}}$

$= -2.303 \times 8.314\times 300\times \log{{10}^{-10}}$

$= 57.441 kJ$

$\Delta H = \Delta G + T\Delta S = 57.441 + 300\times 5\times {10}^{-3}$

$= 58.941 kJ$

$\Delta H = \Delta U + \Delta {n}_{g} RT$
or

$58.941 = \Delta U + \left(-2\right)\times 8.314\times 300\times {10}^{-3}$

$\therefore \Delta U = 63.93 kJ$

Hence, option B is correct.

1.0 g magnesium atoms in vapour phase absorbs 50.0 kJ of energy to convert all Mg into Mg ions. The energy absorbed is needed for the following changes :

$Mg(g)\longrightarrow Mg^+ (g)+ e; \: \Delta H = 740 \: kJ \: mol^{-1}$

$Mg^+(g)\longrightarrow Mg^{2+} (g)+ e; \: \Delta H = 1450 \: kJ \: mol^{-1}$

Find out the % of

$Mg^+$ and

$Mg^{2+}$ in the final mixture.

0%
% $Mg^+ = 68.28$ % , % $of Mg^{2+} = 31.72$ %
0%
% $Mg^+ =58.28$ % , % $of Mg^{2+} = 41.72$ %
0%
% $Mg^+ = 78.28$ % , % $of Mg^{2+} = 21.72$ %
0%
None of these

Explanation

$\displaystyle Moles \: of \: Mg =\frac{1}{24}$

Let

$a$ mole of

$Mg^+$ and

$b$ mole of

$Mg^{2+}$ are present, then

$\displaystyle a+b=\frac{1}{24}$ ... (i)

Also,

$50=a\times 740+ b \times 2190$ ... (ii)

$\therefore$ From eqs. (i) and (ii),

$a=2.845\times 10^{-2}$

$b=1.322\times 10^{-2}$

$of Mg^+ = 68.28\%$

$of Mg^{2+} = 31.72\%$

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 9 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 9 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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