MCQ Questions for Class 11 Medical Chemistry Thermodynamics Quiz 12

Choose the reaction in which

$\Delta H$ is not equal to

$\Delta U$ .

0%
$C\left( graphite \right) +{ O }_{ 2 }\left( g \right) \longrightarrow C{ O }_{ 2 }\left( g \right)$
0%
${ C }_{ 2 }{ H }_{ 4 }\left( g \right) +{ H }_{ 2 }\left( g \right) \longrightarrow { C }_{ 2 }{ H }_{ 6 }\left( g \right)$
0%
$2C\left( graphite \right) +{ H }_{ 2 }\left( g \right) \longrightarrow { C }_{ 2 }{ H }_{ 2 }\left( g \right)$
0%
${ H }_{ 2 }\left( g \right) +{ I }_{ 2 }\left( g \right) \longrightarrow 2HI\left( g \right)$
0%
${ N }_{ 2 }\left( g \right) +{ O }_{ 2 }\left( g \right) \longrightarrow 2NO\left( g \right)$

Explanation

The reaction for which

$\Delta { n }_{ g }\neq 0$ ,

$\Delta H\neq \Delta U$ because

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

$\Delta { n }_{ g }={ n }_{ P }-{ n }_{ R }$
(where,

${ n }_{ P }$ and

${ n }_{ R }$ represents the number of moles of gaseous products and reactants respectively.)

(a)

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

$\Delta { n }_{ g }={ n }_{ P }-{ n }_{ R }=1-1=0$

(b)

${ C }_{ 2 }{ H }_{ 4 }\left( g \right) +{ H }_{ 2 }\left( g \right) \longrightarrow { C }_{ 2 }{ H }_{ 6 }\left( g \right) ;$

$\Delta { n }_{ g }=1-2=-1$

(c)

$2C\left( graphite \right) +{ H }_{ 2 }\left( g \right) \longrightarrow { C }_{ 2 }{ H }_{ 2 }\left( g \right) ;$

$\Delta { n }_{ g }=1-1=0$

(d)

${ H }_{ 2 }\left( g \right) +{ I }_{ 2 }\left( g \right) \longrightarrow 2HI\left( g \right) ;$

$\Delta { n }_{ g }=2-2=0$

(e)

${ N }_{ 2 }\left( g \right) +{ O }_{ 2 }\left( g \right) \longrightarrow 2NO\left( g \right) ;$

$\Delta { n }_{ g }=2-2=0$

Thus, for the reaction given in option (b),

$\Delta H\neq \Delta U$ .

When the total cell emf of a volatic cell is greater than zero, which of the following is true about the reaction quotient

$Q$ and free energy change

$\triangle G$ for the cell reaction?

0%
$Q$ is less than one and $\triangle G$ is greater than zero
0%
$Q$ is greater than one and $\triangle G$ is greater than zero
0%
$Q$ is less than one and $\triangle G$ is less than zero
0%
$Q$ is zero and $\triangle G$ is greater than zero
0%
$Q$ is greater than one and $\triangle G$ is less than zero

Explanation

We know,

$\triangle G = \triangle G^{\circ} + RT \ ln\ Q$

$\because \triangle G^{\circ} = -nFE^{\circ}$

and

$E^{\circ} > 0$

$\therefore \triangle G^{\circ} < 0$

$\triangle G - RT \ ln \ Q < 0$

It is possible when

$Q > 1$ , and

$\triangle G < 0$ .

The enthalpy of combustion for the

$H_{2}$ , cyclohexene and cyclohexane are

$-241, -3800$ and

$-3920kJ\ mol^{-1}$ , respectively. Heat of hydrogenation of cyclohexene is:

0%
$+121 kJ\ mol^{-1}$
0%
$-121kJ\ mol^{-1}$
0%
$+242kJ\ mol^{-1}$
0%
$-242kJ\ mol^{-1}$

Explanation

Different combustion reactions are shown below.

$\displaystyle C_6H_{12}+9O_2 \rightarrow 6CO_2+6H_2O$

$\displaystyle \Delta H_1 = -3920$ kJ/mol ......(1)

$\displaystyle C_6H_{10}+8.5O_2 \rightarrow 6CO_2+5H_2O$

$\displaystyle \Delta H_2 = -3800$ kJ/mol ......(2)

$\displaystyle H_{2}+O_2 \rightarrow H_2O$

$\displaystyle \Delta H_3 = -241$ kJ/mol ......(3)

The balanced chemical equation for the hydrogenation of cyclohexene is

Cyclohexene

$\displaystyle + H_2 \rightarrow$ Cyclohexane .....(4)

Add reaction (3) to reaction (2). Then substract reaction (1) to obtain reaction (4).

$\displaystyle \Delta H_{4} = \Delta H_2- \Delta H_1+ \Delta H_3$

$\displaystyle \Delta H_4=-3800-(-3920)+-241$

$\displaystyle \Delta H_4=-121$ kJ/mol

Hence, the heat of hydrogenation of cyclohexene is

$\displaystyle \Delta H=-121$ kJ/mol

$\Delta H_{f}(H_{2}O)$ is

$-285.20\ kJ\ mol^{-1}$ , then

$\Delta H_{f}^{\circ}(OH^{-})$ is:

0%
$-227.90\ kJ\ mol^{-1}$
0%
$+228.88\ kJ\ mol^{-1}$
0%
$-343.52\ kJ\ mol^{-1}$
0%
$+343.52\ kJ\ mol^{-1}$

Explanation

$H^{+} + OH^{-}\rightarrow H_{2}O, \Delta H = -57.3\ kJ\ mol^{-1}$ (standard value)

$\Delta H = \Delta H_{f}^{\circ} (H_{2}O) - \Delta H_{f}^{\circ} (H^{+}) - \Delta H_{f}^{\circ}(OH^{-})$

$-57.3 = -285.2 - 0 - \Delta H_{f}^{\circ}(OH^{-})$

$\Delta H_{f}^{\circ} (OH^{-1}) = -227.9\ kJ\ mol^{-1}$ .

The densities of graphite and diamond at

$298K$ are

$2.25$ and

$3.31$

$g\ { cm }^{ -3 }$ respectively. If the standard free energy difference is

$1895\ J\ {mol}^{-1}$ , the pressure at which graphite will be transformed into diamond is:

0%
$9.92\times { 10 }^{ 8 }Pa$
0%
$9.92\times { 10 }^{ 7 }Pa$
0%
$9.92\times { 10 }^{ 6 }Pa$
0%
None of these

Explanation

$\quad \Delta G={ W }_{ compress }=-p\Delta V$

$1895=-p\left[ \left( \cfrac { 12 }{ 3.31 } -\cfrac { 12 }{ 2.25} \right) \times { 10 }^{ -6 }{ m }^{ 3 } \right]$

$1895=-p(3.62-5.33)\times { 10 }^{ -6 }$

$p=\cfrac { 1895\times { 10 }^{ 6 } }{ 1.71 } Pa=11.02\times { 10 }^{ 8 }Pa\quad$

At 298 K the entropy of rhombic sulphur 32.04 J/mol K and that of monoclinic sulphur is 32.68 J/mol K. The heat of their combustion are respectively

$-298246$ and

$-297948 \,J$

$mol^{-1}.$

$\Delta \,G$ for the reaction;

$S_{rhormbic} \rightarrow S_{monoclinic}$ will be :

0%
107.28 J
0%
10.728 J
0%
107.28 kJ
0%
10728 J

Explanation

$S_{rhombic}\rightarrow S_{momnoclinic}$

$\Delta H=-297948-(-298246)=298 \, J$

$\Delta S=- 32.68-32.04=0.64\ J/(mol.K)$

$\Delta G=\Delta H-T \Delta S =298 -298 \times 0.64 =107.28\, J$

Given:

$S+\dfrac { 3 }{ 2 } { O }_{ 2 }\longrightarrow S{ O }_{ 3 }+2x$

$kcal$

$S{ O }_{ 2 }+\dfrac { 1 }{ 2 } { O }_{ 2 }\longrightarrow S{ O }_{ 3 }+y$

$kcal$

With the help of the above reactions, find out the heat of formation of

$S{ O }_{ 2 }$ .

0%
$\left( 2x-y \right)$
0%
$\left( x+y \right)$
0%
$\left( 2x+y \right)$
0%
$\left( { 2x }/{ y } \right)$

Explanation

$S+\dfrac { 3 }{ 2 } { O }_{ 2 }\longrightarrow S{ O }_{ 3 }+2x \: kcal$ .....(1)

$S{ O }_{ 2 }+\dfrac { 1 }{ 2 } { O }_{ 2 }\longrightarrow S{ O }_{ 3 }+y \: kcal$ ......(2)

The heat of formation of

$S{ O }_{ 2 }$ can be represented as

$S+\ { O }_{ 2 }\longrightarrow S{ O }_{2 } ; \: \ \ \ \ \ \Delta H_f(SO_2)$ ......(3)

Subtract equation (2) from equation (1) to obtain equation (3), we get-

$\displaystyle \Delta H_f(SO_2)=\left( 2x-y \right)$ kcal

Hence, option A is correct.

A particular reaction at

${27}^{o}C$ for which

$\Delta H > 0$ and

$\Delta S> 0$ is found to be non-spontaneous. The reaction may proceed spontaneously if:

0%
the temperature is decreased
0%
the temperature is kept constant
0%
the temperature is increased
0%
it is carried in open vessel at ${27}^{o}C$

Explanation

Gibb's energy equation:

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

For a spontaneous reaction,

$\Delta G$ should be less than zero.

Therefore, if

$\Delta H>0$ and

$\Delta S>0$ , the

$T\Delta S$ term should be higher than

$\Delta H$ for a spontaneous reaction. So, the temperature should be increased to make the reaction spontaneous.

Although the dissolution of ammonium chloride in water is an endothermic reaction, even then it is spontaneous because:

0%
$\Delta H$ is positive, $\Delta S$ is -ve
0%
$\Delta H$ is +ve, $\Delta S$ is zero
0%
$\Delta H$ is positive, $T\Delta S< \Delta H$
0%
$\Delta H$ is +ve, $\Delta S$ is positive and $\Delta H< T\Delta S$

Explanation

Gibb's free energy equation:

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

For a spontaneous reaction,

$\Delta G<0$

For endothermic reaction,

$\Delta H<0$

For dissolution of ammonium chloride in water,

$\Delta S>0$ . The magnitude of

$T\Delta S$ should be greater than that of

$\Delta H$ so that

$\Delta G<0$ .

Therefore, the dissociation of ammonium chloride in water is an endothermic reaction but is still spontaneous because

$\Delta H$ is

$+ve$ ,

$\Delta S$ is

$+ve$ and

$\Delta H<T\Delta S$ .

A piece of metal weighing

$100$ g is heated to

$80^o$ C and dropped into

$1$ kg of cold water in an insulated container at

$15^o$ C. If the final temperature of the water in the container is

$15.69^o$ C, the specific heat of the metal in

$J/g^o$ .C is:

0%
$0.38$
0%
$0.24$
0%
$0.45$
0%
$0.13$

Explanation

Heat exchanged is given by,

$Q=m\times s\times (T_2-T_1)$ where

$s =$ specific heat

So, when heated metal is dropped into the cold water, we can write:

$m_{metal}\times s_{metal}\times (T_2-T_1)_{metal}=m_{water}\times s_{water}\times (T_2-T_1)_{water}$

$0.1\times s_{metal}\times (80-15.69)=1\times 4.18\times (15.69-15)$

$s_{metal}=0.45 J/g^0\ C$

Hence, option C is correct.

The enthalpy of hydrogenation of benzene is

$-49.8\ kcal/mol$ while its resonance energy is

$36.0\ kcal/mol$ . The enthalpy of formation of benzene is:

0%
$-4.6\ kcal$
0%
$-28.6\ kcal/mol$
0%
$-85.8\ kcal/mol$
0%
$-13.8\ kcal/mol$

The molar heat capacities of Iodine vapour and solid are

$7.8$ and

$14\ cal/mol$ respective enthalpy of sublimation of iodine is

$6096\ cal/mol$ at

$200^{\circ}C$ , then what is

$\Delta H$ the value at

$250^{\circ}C$ in cal/mol.

0%
$5360$
0%
$4740$
0%
$6406$
0%
none of these

Explanation

$\Delta C_p = \dfrac{\Delta H_2 - \Delta H_1}{T_2-T_1}$

$14-7.8 = \Delta {H_2 - 6096}{50}$

$6.2\times 50 = \Delta H_2 - 6096$

$6.2 \times 50 + 6096 = \Delta H_2$

$\Delta H_2 = 310+ 6096 = 6406$ cal/mol

The enthalpy of combustion of methane, graphite and dihydrogen at

$298K$ are

$-890.3$

$kJ{ mol }^{ -1 },-393.5$

$kJ{ mol }^{ -1 },-285.8$

$kJ { mol }^{ -1 }$ respectively. Enthalpy of formation of

${CH}_{4(g)}$ will be:

0%
$-74.8kJ { mol }^{ -1 }$
0%
$-52.27kJ { mol }^{ -1 }$
0%
$+74.8kJ{ mol }^{ -1 }$
0%
$+52.26kJ{ mol }^{ -1 }$

Explanation

${ CH }_{ 4 }+2{ O }_{ 2 }\longrightarrow { CO }_{ 2 }+2{ H }_{ 2 }O;\quad \Delta { H }_{ 1 }=-890kJ\quad { mol }^{ -1 }...(i)\quad$

$C+{ O }_{ 2 }\longrightarrow { CO }_{ 2 };\quad \Delta { H }_{ 2 }=-393.5kJ\quad { mol }^{ -1 }...(ii)$

$2{ H }_{ 2 }+{ O }_{ 2 }\longrightarrow 2{ H }_{ 2 }O;\quad \Delta { H }_{ 3 }=2\times \left( -285.8 \right) kJ\quad { mol }^{ -1 }...(iii)$

Required reaction

$\rightarrow C+2{ H }_{ 2 }\longrightarrow { CH }_{ 4(g) };\quad \Delta { H }_{ f }=?$

From equation

$(ii)$ -

$(i)$ +

$(iii)$

$\Delta { H }_{ f }=\left( -393.5 \right) +890.3+2\left( -285.8 \right) \approx -74.8kJ\quad { mol }^{ -1 }$

Hence option A is correct.

Water is brought to boil under the pressure of

$1.0\ atm$ . When an electric current of

$0.50\ A$ from a

$12\ V$ supply is passed for

$300\ s$ through resistance in thermal contact with it is found that

$0.798\ g$ of water is vapourised. Calculate the molar internal energy change at boiling point

$(373.15\ K)$ .

0%
$37.5kJ\ { mol }^{ -1 }$
0%
$3.75kJ\ { mol }^{ -1 }$
0%
$42.6kJ\ { mol }^{ -1 }$
0%
$4.26kJ\ { mol }^{ -1 }$

Explanation

$\Delta H=$ work done

$=i\times V\times t$

$=0.5\times 12\times 300=1.8\ kJ$

Molar enthalpy of vaporisation

$\Delta { H }_{ m }=\cfrac { \Delta H }{ moles\quad of\quad { H }_{ 2 }O\quad } =\cfrac { \Delta H }{ { n }_{ { H }_{ 2 }O } }$

$=\cfrac { 1.8kJ }{ \cfrac { 0.798 }{ 18 } } =40.6\ kJ{ mol }^{ -1 }$

$\Delta { H }_{ m }=\Delta { E }_{ m }+p\Delta V$

$\Delta { H }_{ m }=\Delta { E }_{ m }+\Delta { n }_{ g }RT\quad$

$\Delta { H }_{ m }=\Delta { E }_{ m }+RT\quad$

[

$\therefore \Delta { n }_{ g }=1\quad for\quad { H }_{ 2 }O(l)\rightleftharpoons { H }_{ 2 }O(g)$ ]

$\therefore$ Molar internal energy change,

$\Delta { E }_{ m }=\Delta { H }_{ m }-RT$

$\Delta { E }_{ m }=40.6-8.314\times { 10 }^{ -3 }\times 373.15=37.5\ kJ { mol }^{ -1 }$

Calculate

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

$UBr_4$ from

$\Delta G^0$ of reaction and the

$S^0$ values.

$U(s)+2Br_2(l)\rightarrow UBr_4(s) ; \, \Delta G^0=-788.6 KJ; \, S^0(J/K-mol) 50.3 , 152.3 , 242.6$

0%
-822.1 KJ/mol
0%
-841.2 KJ/mol
0%
-775.6 KJ/mol
0%
-804.3 KJ/mol

Explanation

$\Delta S^o=(S^o)_{products}-(S^o)_{reactants}$

$\implies \Delta S^o=(242.6)-(1\times 50.3+2\times 152.3)$

$\implies \Delta S^o=-112.3\times 10^{-3}$ $kJ/k-mol$

$T=298$ $K$ (Room Temperature)

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

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

$\implies \Delta H=-788.6+(298)(-112.3\times 10^{-3})$

$\implies \Delta H=-755.1$ $kJ/mol$

$\Delta H=(\Delta H)_{product}-(\Delta H)_{reactant}$

$\Delta H=\Delta H_f^o(UBr_4)-0$

$\therefore \Delta H_f^o=-755.1$ $kJ/mol$

For the reaction,

${ C }_{ 3 }{ H }_{ 8 }\left( g \right) +5{ O }_{ 2 }\left( g \right) \longrightarrow 3C{ O }_{ 2 }\left( g \right) +4{ H }_{ 2 }O\left( l \right)$ at constant temperature,

$\Delta H-\Delta E$ is:

0%
$+3 RT$
0%
$-RT$
0%
$+RT$
0%
$-3RT$

Explanation

${ C }_{ 3 }{ H }_{ 8 }\left( g \right) +5{ O }_{ 2 }\left( g \right) \longrightarrow 3C{ O }_{ 2 }\left( g \right) +4{ H }_{ 2 }O\left( l \right)$

$\displaystyle \Delta n$ is the difference in the number of moles of gaseous products and gaseous reactants.

$\displaystyle \Delta n = (3)-(1+5)=-3$

$\displaystyle \Delta H=\Delta E + \Delta nRT$

$\displaystyle \Delta H-\Delta E = \Delta nRT$

$\displaystyle \Delta H-\Delta E = -3RT$

Hence.option D is correct.

If for

$H_{2}(g) + \dfrac {1}{2} O_{2}(g) \rightarrow H_{2}O(g); \triangle H_{1}$ is the enthalpy of reaction and for

$H_{2}(g) + \dfrac {1}{2}O_{2}(g) \rightarrow H_{2}O(l); \triangle H_{2}$ is enthalpy of reaction then:

0%
$\triangle H_{1} > \triangle H_{2}$
0%
$\triangle H_{1} = \triangle H_{2}$
0%
$\triangle H_{1} < \triangle H_{2}$
0%
$\triangle H_{1} + \triangle H_{2} = 0$

Explanation

In the process,

$H_{2}(g) + \dfrac {1}{2}O_{2}(g)\rightarrow H_{2}O(l)$ ; heat is evolved because the state is changing from gaseous to liquid, i.e., the process is more exothermic because of gas-phase changing to the liquid phase.

Hence,

$\triangle H_{1} < \triangle H_{2}$ .

Ellingham's diagram represents

0%
Change of $\Delta G$ with temperature
0%
Change of $\Delta H$ with temperature
0%
Change of $\Delta G$ with pressure
0%
Change of $\left( \Delta G-T\Delta S \right)$ with temperature

Explanation

Ellingham diagram represents change of

$\Delta G$ with temperature.

$1$ mole of ice at

$0^{\circ}C$ and

$4.6$ mm Hg pressure is converted to water vapour at a constant temperature and pressure. Find

$\Delta H$ if the latent heat of fusion of ice is

$80$ Cal/gm and latent heat of vaporization of liquid water at

$0^{\circ}C$ is

$596$ Cal per gram and the volume of ice in comparison to that of water (vapour) is neglected.

0%
14.53 Kcal/mol
0%
12.16 Kcal/mol
0%
10.22 Kcal/mol
0%
15.62 Kcal/mol

Explanation

$\text{ice} \longrightarrow\text{vapour}$

Given that:-

$\Delta{{H}_{f}} = 80 \; {cal}/{gm} = 80 \times 18 = 1.44 \; {kcal}/{mol}$

$\Delta{{H}_{v}} = 596 \; {cal}/{gm} = 596 \times 18 = 10.728 \; {Kcal}/{mol}$

$\Delta{H} = \Delta{{H}_{f}} + \Delta{{H}_{v}}$

$\Delta{H} = 1.44 + 10.728 = 12.168 \; {Kcal}/{mol}$

Hence, the correct option is

$\text{B}$

$100\ ml\ 0.5 N\ H_{2}SO_{4}$ (strong acid) is neutralised with

$200\ ml\ 0.2\ M\ NH_{4}OH$ in a constant pressure Calorimeter which results in temperature rise of

$1.4^{\circ}C$ . If heat capacity of Calorimeter content is

$1.5\ kJ/^{\circ}C$ . Which statement is/ are correct:

Given:

$HCl + NaOH\rightarrow NaCl + H_{2}O + 57\ kJ$

$CH_{3}COOH + NH_{4}OH \rightarrow CH_{3}COONH_{4} + H_{2}O + 48.1\ kJ$

0%
enthalpy of neutralisation of $HCl$ v/s $NH_{4}OH$ is $-52.5\ kJ/mol$
0%
enthalpy of dissociation (ionization) of $NH_{4}OH$ is $4.5\ kJ/ mol$
0%
enthalpy of dissociation of $CH_{3}COOH$ is $4.6\ kJ/mol$
0%
$\triangle H$ for $2H_{2}O(l)\rightarrow 2H^{+}(aq.) + 2OH^{-}(aq.)$ is $114\ kJ$

Fruend and Long were 2 scientists interested in adsorption.
Once in a discussion Fruend asked "I can see adsorption is spontaneous, but why it is always exothermic?"
Long said"_________________"
Choose Long's answer :

0%
$\triangle$ G > 0 & $\triangle$ S < 0 $\Rightarrow$ $\triangle$ H < 0
0%
$\triangle$ G < 0 & $\triangle$ S < 0 $\Rightarrow$ $\triangle$ H < 0
0%
$\triangle$ G > 0 & $\triangle$ S > 0 $\Rightarrow$ $\triangle$ H > 0
0%
$\triangle$ G , 0 & $\triangle$ S > 0 $\Rightarrow$ $\triangle$ H > 0

Explanation

For the adsorption-

$\Delta G, \Delta H , \Delta S$ all are negative.

According to Gibbs free energy-

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

$\Delta G =\Delta H +T\Delta S$ , as the

$\Delta S = -ve$

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

As the

$\Delta G$ is already negative so

$\Delta H$ also comes negative.

Calculate the heat needed to raise the temperature of

$20g$ iron from

${25}^{o}C$ to

${500}^{o}C$ , if specific heat capacity of iron is

$0.45J{K}^{-1}$

${g}^{-1}$ .

0%
$6732J$
0%
$225J$
0%
$15.66J$
0%
$2250J$

Explanation

Formula for specific heat-

$Q = m C\Delta T$

$Q = .45\times 20\times (475+273) = 6732J$

For the reaction taking place at certain temperature

$NH_{2}COONH_{4}(s)\rightleftharpoons 2NH_{3}(g)$ if equilibrium pressure is

$3X$ bar then

$\Delta G^{\circ}$ would be:

0%
$-RT\ ln\ 9 -3RT\ ln\ X$
0%
$RT\ ln\ 4 -3RT\ ln\ X$
0%
$-3RT\ ln\ 4X$
0%
none of these

Explanation

$NH_2COONH_4 \rightarrow 2NH_3 + CO_2$

$2X$ +

$X$ =

$3X$

$K = {2X}^2\times {X} = {4X}^3$

Standard gibbs free energy

$= -RT\times lnK$

$\Delta G^{o} = - RT \times ln {4X}^3$

$\Delta G^{o} = - 3 RT ln {4X}$

A monoatomic ideal gas undergoes a process in which the ratio of

$P$ to

$V$ at any instant is constant and equals to

$1$ . What is the molar heat capacity of the gas?

0%
$4R/2$
0%
$3R/2$
0%
$5R/2$
0%
$0$

Explanation

Hint: Elements that are stable as single atoms are known as monatomic or monoatomic elements.

Explanation

It is given in the question, $P=V$ and $\frac{P}{V}=1$ ,

Thus,

$dg={{c}_{v}}dt+Pdv$

For $1$ mole of gas $PV=Rt$ ,

$Pdv+VdP=RdT$

$2PdV=RdT$

$PdV=\frac{RdT}{2}$

$dg={{C}_{r}}dt+\frac{RdT}{2}$

$=\frac{3}{2}R+\frac{R}{2}$

$=\frac{4R}{2}$

Correct Option: $A$

Which statement is correct?

0%
$\left (\dfrac {dH}{dT}\right )_{P} < \left (\dfrac {dE}{dT}\right )_{V}$
0%
$\left (\dfrac {dH}{dT}\right )_{P} + \left (\dfrac {dE}{dT}\right )_{V} = R$
0%
$\left (\dfrac {dE}{dV}\right )_{T}$ of ideal gas is zero
0%
All of these

Calculate the hear needed to raise the temperature of

$20g$ from

${25}^{o}C$ to

${500}^{o}C$ , if specific heat capacity of iron is

$0.45J{K}^{-1}$

${g}^{-1}$ .

0%
$4274J$
0%
$225J$
0%
$15.66J$
0%
$2250J$

A solid material supplied with heat at a constant rate. The temperature of material is changing with heat input as shown in figure. What does slope DE represents?

0%
latent heat of liquid
0%
latent heat of vapour
0%
heat capacity of vapour
0%
inverse of heat capacity of vapour

The Gibb's energy for the decomposition of

${Al}_{2}{O}_{3}$ at

${500}^{o}C$ is as follows:

$\cfrac { 2 }{ 3 } { Al }_{ 2 }{ O }_{ 3 }\longrightarrow \dfrac{ 4}{ 3}Al+{ O }_{ 2 }\ ;$

$\Delta G=+960\ kJ$ .

The potential difference needed for the electrolytic reduction of aluminium oxide (

${Al}_{2}{O}_{3}$ ) at

${500}^{o}C$ is at least:

0%
$4.5V$
0%
$3.0V$
0%
$2.5V$
0%
$5.0V$

Explanation

Valancy factor (n) for the given reaction is

$n=4$

So, according to Gibbs free energy-

$\Delta G = -nFE$

$+960\times 10^{3} = -4\times 96500\times E$

$E = -2.48\ V$

So potential difference is

$\approx 2.5V$

$H_2(g) + Cl_2 (g) = 2HCl (g)$ ;

$\Delta{H}$ (298K)

$= -22.06$ kcal. For this reaction,

$\Delta{U}$ is equal to:

0%
$-22.06 + 2\times 10^{-3} \times 298 \times 2 kcal$
0%
$-22.06 + 2 \times 298\ kcal$
0%
$-22.06 - 2 \times 298 \times 4\ kcal$
0%
$-22.06\ kcal$

Explanation

Enthalpy of reaction=

$\Delta H = \Delta E + n_gR\ times T$

$\Delta E = \Delta H - n_gR\times T$

$\Delta E = -22.06- 0\times RT = -22.06Kcal$

Calculate the heat needed to raise the temperature of

$20\ g$ iron from

$25^{\circ}C$ to

$500^{\circ}C$ , if specific heat capacity of iron is

$0.45\ JK^{-1}g^{-1}$ .

0%
$15.66\ J$
0%
$4275\ J$
0%
$225\ J$
0%
$2250\ J$

$9.2g$ of toulene

${C}_{2}{H}_{8}(l)$ is completely burnt in air. The difference in heat change at constant pressure and constant volume at

${27}^{o}C$ is

0%
$-2.5kJ$
0%
$+2.5kJ$
0%
Zero
0%
$-0.50kJ$

A vessel contains

$100$ litres of a liquid X. Heat is supplied to the liquid in such a fashion that, heat given equals change in enthalpy. The volume of the liquid increases by

$2$ litres. If the external pressure is one atm, and

$202.6$ Joules of heat were supplied, then

$[$ U

$-$ total internal energy

$]$ :

0%
$\Delta U = 0, \Delta H=0$
0%
$\Delta U = +202.6J, \Delta H = +202.6$ J
0%
$\Delta U = -202.6J, \Delta H = -202.6$ J
0%
$\Delta U = 0, \Delta H = +202.6J$

$H^{+}(aq) + NaOH (aq) \to Na^{+} + H_{2}O(l)\ \triangle H_{1} = -1390\ cals$

$HCN(aq) + NaOH(aq) \rightarrow Na^{+} + CN^- + H_{2}O(l)\ \triangle H_{2} = -2900\ cals$ .

What is

$\triangle H$ value for

$HCN (aq) \to H^{+}(aq) + CN^{-}(aq)$ ?

0%
$11400\ cals$
0%
$10790\ cals$
0%
$12500\ cals$
0%
$9800\ cals$

Ethyl chloride

$\left( { C }_{ 2 }{ H }_{ 5 }Cl \right)$ , is prepared by reaction of ethylene with hydrogen chloride :

${ C }_{ 2 }{ H }_{ 4 }\left( g \right) +HCl\left( g \right) \longrightarrow { C }_{ 2 }{ H }_{ 5 }Cl\left( g \right) ; \triangle H=-72.3kJ/mol$
What is the value of

$\triangle U$ (in kJ), if

$70\ g$ of ethylene and

$73\ g$ of

$HCL$ are allowed to react at

$300\ K$ ?

0%
$-69.8$
0%
$-180.75$
0%
$-174.5$
0%
$-139.6$

Explanation

Mass of ethylene

$= 70 \; g$

Molar mass of ethylene

$= 28 \; g$

$\therefore$ No. of moles of ethylene

$= \cfrac{70}{28} = 2.5 \text{ moles}$

Mass of

$HCl = 73 \; g$

Molecular mass of

$HCl = 36.5 \; g$

No. of moles of

$HCl = \cfrac{73}{36.5} = 2 \text{ moles}$

Since no. of moles of

$HCl$ are less than that of ethylene.

Hence

$HCl$ will be limiting reagent.

${{C}_{2}{H}_{4}}_{\left( g \right)} + {HCl}_{\left( g \right)} \longrightarrow {{C}_{2}{H}_{5}Cl}_{\left( g \right)}$

From the reaction-

$1$ mole of

$HCl$ reacts with

$1$ mole of

${C}_{2}{H}_{4}$ and gives

$1$ moles of

${C}_{2}{H}_{5}Cl$ .

$\therefore 2$ moles of

$HCl$ will react with

$2$ moles of

${C}_{2}{H}_{4}$ and will give

$2$ mole of

${C}_{2}{H}_{5}Cl$ .

Given that

$\Delta{H} = -72.3 \; {KJ}/{mol} = -72300 \; {KJ}/{mol}$

Given

$\Delta{H}$ is for

$1$ mole of product.

For

$2$ moles,

$\Delta{H} = 2 \times \left( -72300 \right) = 144600 \; {KJ}/{mol}$

$\Delta{{n}_{g}} = {n}_{P} - {n}_{R} = \left( 2 - \left( 2 + 2 \right) \right) = -2$

As we know that,

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

$\Rightarrow \Delta{U} = \Delta{H} - \Delta{{n}_{g}} RT$

$\Delta{U} = -144600 - \left( -2 \times 8.314 \times 300 \right) = -139.61 \; {KJ}/{mol}$ .

For a certain reaction the change in enthalpy and change in entropy are 40.63 kJ

$mol^{-1}$ and 100

$JK^{-1}$ . What is the value of

$\Delta$ at

$27^0C$ and indicate whether the reaction is spontaneous or not

0%
$+10630 J\, mol^{-1}$ ; spontaneous
0%
$+10630 J\, mol^{-1}$ ; non spontaneous
0%
$-7990\, J\, mol^{-1}$ ; spontaneous
0%
$+7900\, J\,mol^{-1}$ ; spontaneous

Stearic acid

$\left[ C{ H }_{ 3 }\ { \left( C{ H }_{ 2 } \right) }_{ 16 }\ C{ O }_{ 2 }H \right]$ is a fatty acid, the part of fat that stores most of the energy.

$1.0\ g$ of stearic acid was burned in a bomb calorimeter. The bomb had a heat capacity of

$652\ J/ C$ . If the temperature of

$500\ g$ water

$\left( c=4.18J/g\ C \right)$ rose from

$25.0$ to

$39.3 C$ , how much heat was released when the strearic acid burned?

$[Given\ { C }_{ p }\left( { H }_{ 0 }O \right) =4.18\ { J }/{ g }\ C]$

0%
$39.21\ kJ$
0%
$29.91\ kJ$
0%
$108\ kJ$
0%
$9.32\ kJ$

For which of the following reaction

$\triangle H \neq \triangle U$ :

0%
$H_2(g) + Cl_2(g)\rightarrow 2HCl(g)$
0%
$N_2(g) + 3H_2(g)\rightarrow 2NH_3(g)$
0%
$NH_4HS_{(s)}\rightarrow NH_3(g) + H_2S(g)$
0%
$CaCO_3(s)\rightarrow CaCO(s) + CO_2(g)$

For Pt,

${H}_{2}$ (

${p}_{1}atm$ )

$\left| { H }^{ \oplus}\left( 1M \right) \right| { H }_{ 2 }\left( { p }_{ 2 }\ atm \right)$ , Pt (where,

${p}_{1}$ and

${p}_{2}$ are pressures) cell reaction will be spontaneous if?

0%
${p}_{1}={p}_{2}$
0%
${p}_{1}> {p}_{2}$
0%
${p}_{2}> {p}_{1}$
0%
${p}_{1}=1atm$

For a certain reaction the change in enthalpy and change in entropy are

$40.63\ kJ\ ,ol\ KJ^{-1}$ . What is the value of

$\triangle G$ at

$27^{\circ}C$ and indicate whether the reaction is spontaneous.

0%
$+10630\ J\ mol^{-1}$ ; spontaneous
0%
$+10630\ J\ mol^{-1}$ ; non spontaneous
0%
$-7990\ J\ mol^{-1}$ ; spontaneous
0%
$+7900\ J\ mol^{-1}$ ; spontaneous

For

$Ag, \overline {C_{P}} (JK^{-1}mol^{-1})$ is given by

$24 + 0.006\ T$ . If temperature of

$3\ mol$ of silver is raised from

$27^{\circ}C$ to its melting point

$927^{\circ}C$ under

$1$ atm pressure then

$\triangle H$ is equal to:

0%
$52\ kJ$
0%
$76.95\ kJ$
0%
$89.62\ kJ$
0%
$38.62\ kJ$

What is free energy change

$\left( \Delta G \right)$ when

$1.0$ mole of water at

${ 100 }^{ o }C$ and

$1$ atm pressure is converted into steam at

${ 100 }^{ o }C$ and

$2\ atm$ pressure?

0%
$0\ cal$
0%
$540\ cal$
0%
$515.4\ cal$
0%
$None\ of\ these$

Explanation

Since, the reaction is being carried out at constant temperature $(100^0C)$ , therefore the reaction is in thermal equilibrium. At equilibrium the value of $\Delta{G}$ is zero.
This can also be proved mathematically

$\Delta{G} = \Delta{G}^0 + RTlnQ$

At equilibrium,

$\Delta{G}^0 = - RTlnK_{eq}$ and

$K_{eq} = Q$

(where

$Q$ is the ratio of initial activity of products to reactants, and

$K_{eq}$ is the ratio of concentrations of products to reactants at equilibrium)

Thus,

$\Delta{G} = -RTlnQ + RTlnQ = 0$

The heat capacity of a bomb calorimeter is

$300\ J/K$ . When

$0.16\ gm$ of methane was burnt in this calorimeter the temperature rose by

$3^{\circ}C$ . The value of

$\triangle U$ per mole will be

0%
$100\ kJ$
0%
$90\ kJ$
0%
$900\ kJ$
0%
$48\ kJ$

$NX$ is produced by the following step of reactions:

$M+{ X }_{ 2 }\longrightarrow M{ X }_{ 2 }$

$3M{ X }_{ 2 }+{ X }_{ 2 }\longrightarrow { M }_{ 3 }{ X }_{ 8 }$

${ M }_{ 3 }{ X }_{ 8 }+{ N }_{ 2 }{ CO }_{ 3 }\longrightarrow NX+{ CO }_{ 2 }+{ M }_{ 3 }{ O }_{ 4 }$

How much

$M$ (metal) is consumed to produce

$206gm$ of

$NX$ ?

(Take atomic weight of

$M=56,N=23,X=80$ )

0%
$42gm$
0%
$336gm$
0%
$\cfrac { 14 }{ 3 } gm$
0%
$\cfrac { 7 }{ 4 } gm$

Explanation

$M+X_{2}\rightarrow MX_{2}$

$3MX_{2}+X_{2}\rightarrow M_{3}X_{8}$

$M_{3}X_{8}+N_{2}CO_{3}\rightarrow NX+CO_{2}+M_{3}O_{4}$

Multiplying Equation 1 by Equation 3, and adding all the equations we get

$3M+3X_{2}\rightarrow 3MX_{2}$

$3MX_{2}+X_{2}\rightarrow M_{3}X_{8}$

$M_{3}X_{8}+N_{2}CO_{3}\rightarrow NX+CO_{2}+M_{3}O_{4}$

Resultant equation:-

$3M+4N_{2}+N_{2}CO_{3}\rightarrow NX+CO_{2}+M_{3}O_{4}$

Mass of

$NX=206\,g$

Molar mass of

$NX=23+80=103\,g/mol$

Moles of

$NX=206\,g\times \cfrac{1\,mol}{103\,g}= 2\,mol \,NX$

Moles of metal

$M= 2\, mol\,NX\times \cfrac{3\,mole\,M}{1\, mole\,NX}=6\,mol\,M$

Mass of metal

$M=6\,mol\,M\times \cfrac{56\,g\,M}{1 mole\,M}=336\,g\,M$

Hence, 336g metal is consumed to produce 206g of

$NX$ .

The heat change during the reaction

$24g\ C$ and

$128g\ S$ following the change:

$C + S_{2} \rightarrow CS_{2}; \triangle H = 22K\ cal$ .

0%
$22\ K\ cal$
0%
$11\ K\ cal$
0%
$44\ K\ cal$
0%
$32\ K\ cal$

$100ml$ of

${O}_{2}$ gas diffuses in

$10$ seconds.

$100ml$ of gas

$x$ diffuses in

$t$ seconds. Gas

$x$ and time

$t$ can be respectively:

0%
${H}_{2}, 2.5$ seconds
0%
${SO}_{2}$ , $16$ seconds
0%
$CO$ , $10$ seconds
0%
$He$ , $4$ seconds

Explanation

If the volume of both the gases are same then,

$\cfrac{t_2}{t_1}=\sqrt{\cfrac{M_2}{M_1}}$

GIven :

$t_1=10s$

$M_1=32g$

$t_2=t$

$M_2=M$

Now,

$\cfrac{t}{10}=\sqrt{\cfrac{M}{32}}$

Since two unknowns are present we can substitute the values from the given options and then find the right answer.

For

$M=2\ t=2.5sec$

$\cfrac{2.5}{10}=\sqrt{\cfrac{2}{32}}$

$\cfrac{1}{4}=\sqrt{\cfrac{1}{16}}$

$\cfrac{1}{4}=\cfrac{1}{4}$

LHS=RHS

Hence, Gas x is

$H_2$ and time t is

$2.5seconds$

For a given reaction,

$\Delta H = 35.5 kJ mol^{-1}$ and

$\Delta S = 83.6 kJ mol^{-1}$ . The reaction is spontaneous at: (Assume that

$\Delta H$ and

$\Delta S$ do not vary with temperature)

0%
T > 425 K
0%
all temperatures
0%
T > 298 K
0%
T < 425 K

Explanation

Solution:- (A)

$T > 425 \; K$

The reaction is

$\begin{cases} \text{spontaneous} & \text{if } \Delta{G} < 0 \\ \text{non-spontaneous} & \text{if } \Delta{G} > 0 \end{cases}$

As we know that,

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

$\because$ The reaction is spontaneous.

$\therefore \Delta{G} < 0$

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

$\Rightarrow T > \cfrac{\Delta{H}}{\Delta{S}}$

Given:-

$\Delta{H} = 35.5 \; {KJ}/{mol} = 35500 \; {J}/{mol}$

$\Delta{S} = 83.6 \; {J}/{mol-K}$

$\therefore T > \cfrac{35500}{83.6}$

$\Rightarrow T > 425 \; K$

Hence the reaction will be spontaneous at

$T > 425 \; K$ .

On the basis of the following thermochemical data

${ H }_{ 2 }{ O }_{ (g) }\longrightarrow { H }_{ (aq) }^{ + }+{ OH }_{ (aq) }^{ - };\Delta H=57.32kJ\quad$

${H}_{2(g)}+\cfrac{1}{2}{O}_{2(g)}\longrightarrow{ H }_{ 2 }{ O }_{ (l) };\Delta H=-286.2kJ$
The value of enthalpy of formation of

${OH}^{-}$ ion at

${ 25 }^{ o }C$ is:

0%
$+288.88kJ$
0%
$-343.52kJ$
0%
$-22.88kJ$
0%
$-228.88kJ$

Explanation

Solution:- (D)

$-228.88 \; KJ$

${{H}_{2}O}_{\left( l \right)} \longrightarrow {{H}^{+}}_{\left( aq. \right)} + {{OH}^{-}}_{\left( aq. \right)}; \quad \Delta{H} = 57.32 \; KJ ..... \left( 1 \right)$

${{H}_{2}}_{\left( g \right)} + \cfrac{1}{2} {{O}_{2}}_{\left( g \right)} \longrightarrow {{H}_{2}O}_{\left( l \right)}; \quad \Delta{H} = -286.2 \; KJ ..... \left( 2 \right)$

Adding both the reaction will give enthalpy of formation of

${OH}^{-}$ .

Therefore,

${{H}_{2}O}_{\left( l \right)} + {{H}_{2}}_{\left( g \right)} + \cfrac{1}{2} {{O}_{2}}_{\left( g \right)} \longrightarrow {{H}^{+}}_{\left( aq. \right)} + {{OH}^{-}}_{\left( aq. \right)} + {{H}_{2}O}_{\left( l \right)}; \quad \Delta{H} = -286.2 + 57.32$

${{H}_{2}}_{\left( g \right)} + \cfrac{1}{2} {{O}_{2}}_{\left( g \right)} \longrightarrow {{H}^{+}}_{\left( aq. \right)} + {{OH}^{-}}_{\left( aq. \right)}; \quad \Delta{H} = -228.88 \; KJ$

Hence the enthalpy of formation of

${OH}^{-}$ ion is

$-228.88 \; KJ$ .

Heat of reaction for the equation,

$A(s) + B(g) \rightarrow 2C(g)$ is

$40\ kJ$ at

$300\ K$ at constant volume. Hence, heat of reaction at constant pressure and at

$300\ K$ is___________.

0%
$42.5\ kJ$
0%
$37.5\ kJ$
0%
$40.0\ kJ$
0%
$30.0\ kJ$

If an electron is moving in an orbit with total energy

$U = -\dfrac {e^{2}}{r} r$ is radius of the orbit then find the speed of the electron.

0%
$\dfrac {e}{\sqrt {mr}}$
0%
$e\dfrac {e}{\sqrt {mr}}$
0%
$\dfrac {e}{r}\sqrt {\dfrac {2}{m}}$
0%
$\dfrac {2e}{\sqrt {mr}}$

${E}^{o}$ of an electrode half reaction is related to

$\Delta { G }^{ o }$ by the equation,

${ E }^{ o }=-\Delta { G }^{ o }/nF$ . If the amount of

${Ag}^{+}$ in the half reaction,

${ Ag }^{ + }+{ e }^{ - }\longrightarrow Ag$ is tripled then:

0%
$n$ is tripled
0%
$\Delta { G }^{ o }$ reduces to one third
0%
${E}^{o}$ reduces to one third
0%
None of the above

Explanation

$E^{\circ}$ of an electrode half reaction is related to

$\triangle G^{\circ}$

by the equation.

$E^{\circ}=\cfrac{-\triangle G^{\circ}}{nF}$ . If the amount of

$Ag^{+}$ in half reaction.

$Ag^{+}+e\rightarrow Ag$ is tripled then

$n$ is tripled.

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 12 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 12 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.