Thermodynamics — Page 9 | JEE Chemistry

Q81

If

\quad C

(diamond

) \rightarrow C

(graphite)

+X \mathrm{~kJ} \mathrm{~mol}^{-1}

C (diamond)

+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+\mathrm{Y} \mathrm{kJ} \mathrm{mol}{ }^{-1}

C (graphite)

+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+\mathrm{Z} \mathrm{kJ} \mathrm{mol}^{-1}

at constant temperature. Then

A −X = Y + Z

B X = Y − Z

C X = −Y + Z

D X = Y + Z

Correct Answer

Option B

Solution

Given,

{C_{(diamond)}}\overset{{}}\longrightarrow {C_{(graphite)}} + X\,kJ\,mo{l^{ - 1}}

............ (1)

{C_{(diamond)}} + {O_2}(g)\overset{{}}\longrightarrow C{O_2}(g) + Y\,kJ\,mo{l^{ - 1}}

............. (2)

{C_{(graphite)}} + {O_2}(g)\overset{{}}\longrightarrow C{O_2}(g) + Z\,kJ\,mo{l^{ - 1}}

..............

(3) Condition for temperature : constant Hess's law is applied here.

(Hess's law of constant heat summation) The given reactions (2) and (3), (2) - (3) $\Rightarrow$

{C_{(diamond)}} + {O_2}(g) - \left( {{C_{(graphite)}} + {O_2}(g)} \right)\overset{{}}\longrightarrow C{O_2}(g) + Y - \left( {C{O_2}(g) + Z} \right)

{C_{(diamond)}} + {O_2}(g) - {C_{(grpahite)}} - {O_2}(g)\overset{{}}\longrightarrow C{O_2}(g) + Y - C{O_2}(g) - Z

{C_{(diamond)}} - {C_{(graphite)}} \to Y - Z

{C_{(diamond)}} \to {C_{(graphite)}} + (Y - Z)

............. (4) Comparing (1) and (4),

X = Y - Z

So, the correct answer is option (2)

X = Y - Z

Q82

A liquid when kept inside a thermally insulated closed vessel at

25^{\circ} \mathrm{C}

was mechanically stirred from outside. What will be the correct option for the following thermodynamic parameters ?

A

\Delta \mathrm{U}=0, \mathrm{q}0

B

\Delta \mathrm{U}>0, \mathrm{q}=0, \mathrm{w}>0

C

\Delta \mathrm{U}=0, \mathrm{q}=0, \mathrm{w}=0

D

\Delta \mathrm{U}0

Correct Answer

Option B

Solution

Let’s analyze the situation step by step: System: The liquid inside a thermally insulated (adiabatic) and closed vessel.

Process: The liquid is mechanically stirred from outside.

Heat Exchange ( $q$ ) Because the vessel is thermally insulated, there is no heat exchange between the system and surroundings.

Hence, $q = 0.$ Work ( $w$ ) The external mechanical stirring does work on the system by agitating the liquid.

Work done on the system is conventionally taken as positive: $w > 0.$ Change in Internal Energy ( $\Delta U$ ) From the First Law of Thermodynamics, $\Delta U \;=\; q \;+\; w.$ Since $q = 0$ and $w > 0$ , we get $\Delta U = w > 0.$ Hence: $\boxed{ \Delta U > 0,\quad q = 0,\quad w > 0. }$ Answer: Option B is correct.

Q83

Ice at

-5^{\circ} \mathrm{C}

is heated to become vapor with temperature of

110^{\circ} \mathrm{C}

at atmospheric pressure. The entropy change associated with this process can be obtained from

A

\int_{268 \mathrm{~K}}^{273 \mathrm{~K}} \mathrm{C}_{\mathrm{p}, \mathrm{m}} \mathrm{dT}+\dfrac{\Delta \mathrm{H}_{\mathrm{m}} \text{, fusion }}{\mathrm{T}_{\mathrm{f}}}+\dfrac{\Delta \mathrm{H}_{\mathrm{m}, \text{ vaporisation }}}{\mathrm{T}_{\mathrm{b}}}+\int_{273 \mathrm{~K}}^{373 \mathrm{~K}} \mathrm{C}_{\mathrm{p}, \mathrm{m}} \mathrm{dT}+\int_{373 \mathrm{~K}}^{383 \mathrm{~K}} \mathrm{C}_{\mathrm{p}, \mathrm{m}} \mathrm{dT}

B

\int_{268 \mathrm{~K}}^{383 \mathrm{~K}} \mathrm{C}_{\mathrm{p}} \mathrm{dT}+\dfrac{\Delta \mathrm{H}_{\text{melting }}}{273}+\dfrac{\Delta \mathrm{H}_{\text{boiling }}}{373}

C

\int_{268 \mathrm{~K}}^{383 \mathrm{~K}} \mathrm{C}_{\mathrm{p}} \mathrm{dT}+\dfrac{\mathrm{q}_{\text{rev }}}{\mathrm{T}}

D

\int_{268 \mathrm{~K}}^{273 \mathrm{~K}} \dfrac{\mathrm{C}_{\mathrm{p}, \mathrm{m}}}{\mathrm{T}} \mathrm{dT}+\dfrac{\Delta \mathrm{H}_{\mathrm{m}}, \text{ fusion }}{\mathrm{T}_{\mathrm{f}}}+\dfrac{\Delta \mathrm{H}_{\mathrm{m}, \text{ vaporisation }}^{373 \mathrm{~K}}}{\mathrm{~T}_{\mathrm{b}}}+\int_{273 \mathrm{~K}} \dfrac{\mathrm{C}_{\mathrm{p}, \mathrm{m}} \mathrm{dT}}{T}+\int_{373 \mathrm{~K}}^{383 \mathrm{~K}} \dfrac{\mathrm{C}_{\mathrm{p}, \mathrm{m}} \mathrm{dT}}{\mathrm{T}}

Correct Answer

Option D

Solution

\begin{aligned} & \Delta \mathrm{S}_{\text{overall }}=\Delta \mathrm{S}_1+\Delta \mathrm{S}_2+\Delta \mathrm{S}_3+\Delta \mathrm{S}_4+\Delta \mathrm{S}_5 \\ & \Delta \mathrm{~S}_2=\frac{\Delta \mathrm{H}_{\mathrm{m} \text{ fusion }}}{273} \mathrm{~T}_{\mathrm{f}}=273^{\prime} \mathrm{K}^{\prime} \\ & \Delta \mathrm{S}_3=\int\limits_{273}^{373} \frac{\mathrm{C}_{\mathrm{p}, \mathrm{~m}} \mathrm{dT}}{\mathrm{~T}} \\ & \Delta \mathrm{~S}_4=\frac{\Delta \mathrm{H}_{\mathrm{m} \text{ vaporisation }}}{373} \mathrm{~T}_{\mathrm{b}}=373^{\prime} \mathrm{K}^{\prime} \\ & \Delta \mathrm{S}_5=\int\limits_{373}^{383} \frac{\mathrm{C}_{\mathrm{p}, \mathrm{~m}} \mathrm{dT}}{\mathrm{~T}} \\ & \text{ Answer }=(2) \end{aligned}

Q84

Let us consider an endothermic reaction which is non-spontaneous at the freezing point of water. However, the reaction is spontaneous at boiling point of water. Choose the correct option.

A Both

\Delta \mathrm{H}

and

\Delta \mathrm{S}

are (-ve)

B

\Delta \mathrm{H}

is

(+\mathrm{ve})

but

\Delta \mathrm{S}

is (-ve)

C

\Delta \mathrm{H}

is

(-\mathrm{ve})

but

\Delta \mathrm{S}

is (+ve)

D Both

\Delta \mathrm{H}

and

\Delta \mathrm{S}

are (+ve)

Correct Answer

Option D

Solution

Reaction is spontaneous at relatively high temperature and non-spontaneous at low temperature $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ It is only possible when $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ both are positive.

Option (1)

Q85

\begin{aligned} & \mathrm{S}(\mathrm{~g})+\frac{3}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{SO}_3(\mathrm{~g})+2 x \mathrm{kcal} \\ & \mathrm{SO}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{SO}_3(\mathrm{~g})+y \mathrm{kcal} \end{aligned}

The heat of formation of

\mathrm{SO}_2(\mathrm{~g})

is given by :

A

2 x+y

kcal

B

\dfrac{2 x}{y} \mathrm{~kcal}

C

y-2 x \mathrm{~kcal}

D

x+y \mathrm{~kcal}

Correct Answer

Option C

Solution

To determine the heat of formation for $\mathrm{SO}_2(\mathrm{~g})$ , we need to consider the given reactions and their enthalpy changes.

The reactions are: $\mathrm{S}(\mathrm{~g}) + \dfrac{3}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{SO}_3(\mathrm{~g}) + 2x \mathrm{kcal}$ $\mathrm{SO}_2(\mathrm{~g}) + \dfrac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{SO}_3(\mathrm{~g}) + y \mathrm{kcal}$ For reaction 2, we have: $\Delta \mathrm{H}_{\mathrm{r}} = \left(\Delta \mathrm{H}_{\mathrm{f}}\right)_{\mathrm{SO}_3} - \left(\Delta \mathrm{H}_{\mathrm{f}}\right)_{\mathrm{SO}_2} = -y$ Given that: $-\mathrm{y} = -2 \mathrm{x} - \left(\Delta \mathrm{H}_{\mathrm{f}}\right)_{\mathrm{SO}_2}$ Rearranging this equation gives: $\left(\Delta \mathrm{H}_{\mathrm{f}}\right)_{\mathrm{SO}_2} = y - 2x$ Thus, the heat of formation for $\mathrm{SO}_2(\mathrm{~g})$ is $y - 2x$ kcal.

Q86

Which of the following mixing of 1 M base and 1 M acid leads to the largest increase in temperature?

A 50 mL HCl and 20 mL NaOH

B 30 mL HCl and 30 mL NaOH

C

45 \mathrm{~mL} \mathrm{~CH}_3 \mathrm{COOH}

and 25 mL NaOH

D

30 \mathrm{~mL} \mathrm{~CH}_3 \mathrm{COOH}

and 30 mL NaOH

Correct Answer

Option B

Solution

The rise in temperature of neutralization reaction will be maximum for maximum number of moles of strong acid and strong base neutralized and lower volume of final solution.

Option (D) and (C) have weak acids and in option (A) only 20 mmol of HCl will be neutralized with 70 mL final volume.

Q87

Ice and water are placed in a closed container at a pressure of 1 atm and temperature 273.15 K . If pressure of the system is increased 2 times, keeping temperature constant, then identify correct observation from following

A Liquid phase disappears completely.

B The amount of ice decreases.

C The solid phase (ice) disappears completely.

D Volume of system increases .

Correct Answer

Option C

Solution

If pressure is made two time then mixture of ice and water will completely convert into water (liquid) form.

Q88

Total enthalpy change for freezing of 1 mol of water at

10^{\circ} \mathrm{C}

to ice at

-10^{\circ} \mathrm{C}

is ________ (Given:

\Delta_{\text{fus }} \mathrm{H}=x \mathrm{~kJ} / \mathrm{mol}

\begin{aligned} & \mathrm{C}_{\mathrm{p}}\left[\mathrm{H}_2 \mathrm{O}(\mathrm{l})\right]=y \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \\ & \mathrm{C}_{\mathrm{p}}\left[\mathrm{H}_2 \mathrm{O}(\mathrm{~s})\right]=z \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \end{aligned}

A

-x-10 y-10 z

B

x-10 y-10 z

C

-10(100 x+y+z)

D

10(100 \mathrm{x}+y+z)

Correct Answer

Option C

Solution

\begin{aligned} & \Delta H=1 \times y(0-10)-x \times 1000+1 \times z\left(-10^{\circ}-0^{\circ}\right) \\ & \Delta H=-10(100 x+y+z) \text{ Joule. } \end{aligned}

Q89

Arrange the following in order of magnitude of work done by the system/on the system at constant temperature. (a)

\left|w_{\text{reversible }}\right|

for expansion in infinite stages. (b)

\left|w_{\text{irreversible }}\right|

for expansion in single stage. (c)

\left|\mathrm{w}_{\text{reversible }}\right|

for compression in infinite stages. (d)

\left|w_{\text{irreversible }}\right|

for compression in single stage. Choose the correct answer from the options given below :

A

\mathbf{}_{} \mathrm{d}>\mathrm{c}=\mathrm{a}>\mathrm{b}

B

\mathrm{c}=\mathrm{a}>\mathrm{d}>\mathrm{b}

C

a>c>b>d

D

a>b>c>d

Correct Answer

Option A

Solution

\begin{aligned} &\text{ For isothermal process }\\ &\begin{aligned} & \left|\mathrm{W}_{\text{reversible }}\right|_{\text{expansion }}=\left|\mathrm{W}_{\text{reversible }}\right|_{\text{compression }} \\ & =-\mathrm{nRT} \ln \frac{\mathrm{~V}_{\mathrm{f}}}{\mathrm{~V}_{\mathrm{i}}} \\ & \left|\mathrm{~W}_{\text{irreversible }}\right|_{\text{expansion }}<\left|\mathrm{W}_{\text{irreversible }}\right|_{\text{compression }} \\ & \mathrm{d}>\mathrm{c}=\mathrm{a}>\mathrm{b} \\ & \left|\mathrm{~W}_{\text{irreversible }}\right|_{\text{expansion }}=-\mathrm{P}_{\text{ext }}\left(\mathrm{V}_{\mathrm{f}}-\mathrm{V}_{\mathrm{i}}\right) \end{aligned} \end{aligned}

Graphical representation We can compare work by area of PV graph.

Q90

Consider the given data : (a)

\mathrm{HCl}(\mathrm{g})+10 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl} .10 \mathrm{H}_2 \mathrm{O} \Delta \mathrm{H}=-69.01 \mathrm{~kJ} \mathrm{~mol}^{-1}

(b)

\mathrm{HCl}(\mathrm{g})+40 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl} .40 \mathrm{H}_2 \mathrm{O} \Delta \mathrm{H}=-72.79 \mathrm{~kJ} \mathrm{~mol}^{-1}

Choose the correct statement :

A The heat of dilution for the

\mathrm{HCl}\left(\mathrm{HCl} .10 \mathrm{H}_2 \mathrm{O}\right.

to

\left.\mathrm{HCl} .40 \mathrm{H}_2 \mathrm{O}\right)

is

3.78 \mathrm{~kJ} \mathrm{~mol}^{-1}

.

B Dissolution of gas in water is an endothermic process.

C The heat of solution depends on the amount of solvent.

D The heat of formation of HCl solution is represented by both (a) and (b).

Correct Answer

Option C

Solution

From the given information $\Delta \mathrm{H}$ is negative so it means dissolution of gas $\mathrm{HCl}(\mathrm{g})$ is exothermic.

\mathrm{HCl}(\mathrm{~g})+10 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl} .10 \mathrm{H}_2 \mathrm{O}\quad\text{..... (1)}

\begin{aligned} & \Delta \mathrm{H}_1=-69.01 \frac{\mathrm{~kJ}}{\mathrm{~mol}} \\ & \mathrm{HCl}(\mathrm{~g})+40 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl} .40 \mathrm{H}_2 \mathrm{O}\quad\text{..... (2)} \end{aligned}

\Delta \mathrm{H}_2=-72.79 \frac{\mathrm{~kJ}}{\mathrm{~mol}}

Hence heat of solution depends upon amount of solvent By equation ............. (2) $-$ equation ........... (1)

\mathrm{HCl} .10 \mathrm{H}_2 \mathrm{O}+30 \mathrm{H}_2 \mathrm{O}(\ell) \rightarrow \mathrm{HCl} .40 \mathrm{H}_2 \mathrm{O}

So Heat of dilution $=-72.79-(-69.01)$

=-3.78 \frac{\mathrm{~kJ}}{\mathrm{~mol}}

Hence option (3) is incorrect. For heat of formation reactant should be in elemental form hence option (4) is incorrect