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Thermodynamics

NEET Chemistry · 93 questions · Page 1 of 10 · Click an option or "Show Solution" to reveal answer

Q1
Consider the following reaction : 2 A( g)+B( g)→2D( g)ΔU⊖=−10 kJ mol−1 and ΔS⊖=−44JK−1 at 298 K. \begin{aligned} & 2 \mathrm{~A}(\mathrm{~g})+\mathrm{B}(\mathrm{~g}) \rightarrow 2 \mathrm{D}(\mathrm{~g}) \\ & \Delta \mathrm{U}^{\ominus}=-10 \mathrm{~kJ} \mathrm{~mol}^{-1} \text{ and } \Delta \mathrm{S}^{\ominus}=-44 \mathrm{JK}^{-1} \text{ at } 298 \mathrm{~K} . \end{aligned} Identify the correct option with ΔG⊖\Delta \mathrm{G}^{\ominus} for the reaction and spontaneity of the reaction at 298 K . (Given : R=8.31 J mol−1 K−1\mathrm{R}=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} )
A −1.635 kJ mol−1-1.635 \mathrm{~kJ} \mathrm{~mol}^{-1}, spontaneous
B −0.63568 kJ mol−1-0.63568 \mathrm{~kJ} \mathrm{~mol}^{-1}, spontaneous
C +0.63568 kJ mol−1+0.63568 \mathrm{~kJ} \mathrm{~mol}^{-1}, non-spontaneous
D +1.635 kJ mol−1+1.635 \mathrm{~kJ} \mathrm{~mol}^{-1}, non-spontaneous
Correct Answer
Option C
Solution

2 A( g)+B(g)→2D(g)2 \mathrm{~A}(\mathrm{~g})+\mathrm{B}(\mathrm{g}) \rightarrow 2 \mathrm{D}(\mathrm{g})

ΔU∘=−10 kJ/molΔ S∘=−44 J/KΔH∘=ΔU∘+ΔngRTΔH∘=−10−1×298×(8.31)1000=−10−2.48=−12.48 kJ/molΔG∘=ΔH∘−TΔ S∘=−12.48−298×(−44)1000=−12.48+13.112=+0.632 kJ/mol\begin{aligned} \Delta \mathrm{U}^{\circ} & =-10 \mathrm{~kJ} / \mathrm{mol} \\ \Delta \mathrm{~S}^{\circ} & =-44 \mathrm{~J} / \mathrm{K} \\ \Delta \mathrm{H}^{\circ} & =\Delta \mathrm{U}^{\circ}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \\ \Delta \mathrm{H}^{\circ} & =-10-\frac{1 \times 298 \times(8.31)}{1000}=-10-2.48=-12.48 \mathrm{~kJ} / \mathrm{mol} \\ \Delta \mathrm{G}^{\circ} & =\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{~S}^{\circ} \\ & =-12.48-\frac{298 \times(-44)}{1000} \\ & =-12.48+13.112 \\ & =+0.632 \mathrm{~kJ} / \mathrm{mol} \end{aligned}

Since ΔG∘\Delta \mathrm{G}^{\circ} comes out to be positive, so given process is non-spontaneous.

Q2
At a certain temperature, T(K)\mathrm{T}(\mathrm{K}), during a process, 500 J is absorbed by the system and work of 200 J is done by the system. Then change in internal energy of the system is :
A 400 J
B 300 J
C 700 J
D 500 J
Correct Answer
Option B
Solution

From first law of thermodynamics

ΔU=q+wq=+500 Jw=−200 JΔU=500−200=300 J\begin{aligned} \Delta U & =q+w \\ q & =+500 \mathrm{~J} \\ w & =-200 \mathrm{~J} \\ \Delta U & =500-200 \\ & =300 \mathrm{~J} \end{aligned}
Q3
The standard heat of formation, in kcal/mol\mathrm{kcal} / \mathrm{mol} of Ba2+\mathrm{Ba}^{2+} is : [Given : standard heat of formation of SO42−\mathrm{SO}_4^{2-} ion (aq)=−216kcal/mol(\mathrm{aq})=-216 \mathrm{kcal} / \mathrm{mol}, standard heat of crystallisation of BaSO4( s)=−4.5kcal/mol\mathrm{BaSO}_4(\mathrm{~s})=-4.5 \mathrm{kcal} / \mathrm{mol}, standard heat of formation of BaSO4( s)=−349kcal/mol]\left.\mathrm{BaSO}_4(\mathrm{~s})=-349 \mathrm{kcal} / \mathrm{mol}\right]
A +133.0
B +220.5
C −-128.5
D −-133.0
Correct Answer
Option C
Solution

Let's consider the following equations: Formation of SO42−\mathrm{SO}_4^{2-}: S+2O2+2e−⟶SO42−ΔHf=−216kcal/mol \mathrm{S} + 2 \mathrm{O}_2 + 2 \mathrm{e}^{-} \longrightarrow \mathrm{SO}_4^{2-} \quad \Delta \mathrm{H}_{\mathrm{f}} = -216 \mathrm{kcal/mol} Crystallization of BaSO4(s)\mathrm{BaSO}_4(\mathrm{s}): Ba2+(g)+SO42−(g)⟶BaSO4(s)ΔHcrystallisation=−4.5kcal/mol \mathrm{Ba}^{2+}(\mathrm{g}) + \mathrm{SO}_4^{2-}(\mathrm{g}) \longrightarrow \mathrm{BaSO}_4(\mathrm{s}) \quad \Delta \mathrm{H}_{\text{crystallisation}} = -4.5 \mathrm{kcal/mol} Formation of BaSO4(s)\mathrm{BaSO}_4(\mathrm{s}): Ba+S+2O2⟶BaSO4(s)ΔHf(BaSO4)=−349kcal/mol \mathrm{Ba} + \mathrm{S} + 2 \mathrm{O}_2 \longrightarrow \mathrm{BaSO}_4(\mathrm{s}) \quad \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{BaSO}_4) = -349 \mathrm{kcal/mol} We need to find the standard heat of formation for Ba2+ \mathrm{Ba}^{2+} .

To determine this, apply the following relationship derived from the equations: Ba(s)⟶Ba2+(g)+2e− \mathrm{Ba}(\mathrm{s}) \longrightarrow \mathrm{Ba}^{2+}(\mathrm{g}) + 2 \mathrm{e}^{-} Using the data provided: Start from equation (3), then subtract equations (1) and (2): −349−(−4.5)−(−216) -349 - (-4.5) - (-216) Simplifying the expression: −349+4.5+216 -349 + 4.5 + 216 =−349+220.5 = -349 + 220.5 =−128.5 kcal/mol = -128.5 \, \mathrm{kcal/mol} This calculation shows that the standard heat of formation for Ba2+ \mathrm{Ba}^{2+} is −128.5 kcal/mol-128.5 \, \mathrm{kcal/mol}.

Q4
Choose the correct statement for the work done in the expansion and heat absorbed or released when 5 litres of an ideal gas at 10 atmospheric pressure isothermally expands into vacuum until volume is 15 litres :
A Both the heat and work done will be greater than zero
B Heat absorbed will be less than zero and work done will be positive
C Work done will be zero and heat will also be zero
D Work done will be greater than zero and heat will remain zero
Correct Answer
Option C
Solution

Since it is isothermal,

ΔT=0\Delta \mathrm{T}=0
ΔU=nCvΔT=0\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}=0

Since expansion is taking place against vacuum

Pext =0W=−Pext ΔV=0\begin{aligned} & P_{\text{ext }}=0 \\ & W=-P_{\text{ext }} \Delta V=0 \end{aligned}

From first law of thermodynamics,

ΔU=q+W0=q+0q=0\begin{aligned} & \Delta U=q+W \\ & 0=q+0 \\ & q=0 \end{aligned}
Q5
For the following reaction at 300 K300 \mathrm{~K} A2( g)+3 B2( g)→2AB3( g)\mathrm{A}_2(\mathrm{~g})+3 \mathrm{~B}_2(\mathrm{~g}) \rightarrow 2 \mathrm{AB}_3(\mathrm{~g}) the enthalpy change is +15 kJ+15 \mathrm{~kJ}, then the internal energy change is :
A 19988.4 J19988.4 \mathrm{~J}
B 200 J200 \mathrm{~J}
C 1999 J1999 \mathrm{~J}
D 1.9988 kJ1.9988 \mathrm{~kJ}
Correct Answer
Option A
Solution

To determine the internal energy change (ΔU) for the reaction at

300 K300 \mathrm{~K}

, we will use the relationship between enthalpy change (ΔH) and internal energy change:

ΔH=ΔU+ΔnRT\Delta H = \Delta U + \Delta nRT

where

ΔH\Delta H

is the enthalpy change,

ΔU\Delta U

is the internal energy change,

Δn\Delta n

is the change in moles of gas,

RR

is the universal gas constant, and

TT

is the temperature. Given:

ΔH=+15 kJ=15000 J\Delta H = +15 \mathrm{~kJ} = 15000 \mathrm{~J}
T=300 KT = 300 \mathrm{~K}
R=8.314 J mol−1 K−1R = 8.314 \, \mathrm{J \, mol^{-1} \, K^{-1}}
Δn=moles of products−moles of reactants\Delta n = \text{moles of products} - \text{moles of reactants}

From the balanced chemical equation:

A2( g)+3B2( g)→2AB3( g)\mathrm{A}_2(\mathrm{~g}) + 3 \mathrm{B}_2(\mathrm{~g}) \rightarrow 2 \mathrm{AB}_3(\mathrm{~g})

Moles of reactants = 1 (for

A2\mathrm{A}_2

) + 3 (for

B2\mathrm{B}_2

) = 4 moles Moles of products = 2 (for

AB3\mathrm{AB}_3

) Therefore,

Δn\Delta n

= 2 (products) - 4 (reactants) = -2 Now, substituting these values into the equation:

ΔH=ΔU+ΔnRT\Delta H = \Delta U + \Delta nRT
15000=ΔU+(−2)(8.314)(300)15000 = \Delta U + (-2)(8.314)(300)

Simplify the equation:

15000=ΔU−4988.415000 = \Delta U - 4988.4

Therefore:

ΔU=15000+4988.4=19988.4 J\Delta U = 15000 + 4988.4 = 19988.4 \mathrm{~J}

Hence, the internal energy change is: Option A:

19988.4 J19988.4 \mathrm{~J}
Q6
The work done during reversible isothermal expansion of one mole of hydrogen gas at 25∘C25^{\circ} \mathrm{C} from pressure of 20 atmosphere to 10 atmosphere is (Given R=2.0 cal K−1 mol−1\mathrm{R}=2.0 \mathrm{~cal} \mathrm{~K}^{-1} \mathrm{~mol}^{-1})
A 0 calorie
B −-413.14 calories
C 413.14 calories
D 100 calories
Correct Answer
Option B
Solution

The work done in a reversible isothermal process can be calculated using the formula:

W=−nRTln⁡(VfVi)W = -nRT \ln\left(\frac{V_f}{V_i}\right)

Here:

WW

is the work done by the gas.

nn

is the number of moles of the gas.

RR

is the universal gas constant.

TT

is the temperature in Kelvin.

ViV_i

and

VfV_f

are the initial and final volumes of the gas, respectively.

However, since the volumes are not directly provided but the pressures are given, we use the ideal gas law,

PV=nRTPV = nRT

, to relate pressures and volumes at the same temperature and amount of gas: For an ideal gas undergoing a change at constant temperature, we can also write the work done in terms of the initial and final pressures:

W=−nRTln⁡(PiPf)W = -nRT \ln\left(\frac{P_i}{P_f}\right)

where:

PiP_i

and

PfP_f

are the initial and final pressures, respectively. Given:

n=1n = 1

(one mole of hydrogen)

T=25∘C=25+273.15=298.15 KT = 25^{\circ} \mathrm{C} = 25 + 273.15 = 298.15 \, \mathrm{K}
R=2.0 cal K−1 mol−1R = 2.0 \, \mathrm{cal} \, \mathrm{K}^{-1} \, \mathrm{mol}^{-1}
Pi=20 atmP_i = 20 \, \mathrm{atm}
Pf=10 atmP_f = 10 \, \mathrm{atm}

Substituting all values into the formula:

W=−1⋅2.0⋅298.15ln⁡(2010)W = -1 \cdot 2.0 \cdot 298.15 \ln\left(\frac{20}{10}\right)
W=−2.0⋅298.15⋅ln⁡(2)W = -2.0 \cdot 298.15 \cdot \ln(2)

Now, solving this using the value of

ln⁡(2)≈0.693\ln(2) \approx 0.693

:

W≈−2.0⋅298.15⋅0.693W \approx -2.0 \cdot 298.15 \cdot 0.693
W≈−413.14 caloriesW \approx -413.14 \, \text{calories}

Thus, the work done during the process is about

−413.14 calories-413.14 \, \text{calories}

, indicating that this amount of energy was done by the system (expansion work being done by the gas against external pressure), and hence is negative as it indicates work done by the system.

Therefore, the correct option is: Option B:

−413.14 calories-413.14 \, \text{calories}
Q7
Consider the following reaction :- 2H2( g)+O2( g)→2H2O(g)ΔrH∘=−483.64 kJ. 2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \Delta_{\mathrm{r}} \mathrm{H}^{\circ}=-483.64 \mathrm{~kJ} \text{. } What is the enthalpy change for decomposition of one mole of water? (Choose the right option).
A 120.9 kJ
B 241.82 kJ
C 18 kJ
D 100 kJ
Correct Answer
Option B
Solution

Decomposition for 1 mole of water

H2O(g)→H2( g)+12O2( g);ΔH=+483.642ΔH=+241.82 kJ\begin{aligned} & \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) ; \Delta \mathrm{H}=+\frac{483.64}{2} \\ & \Delta \mathrm{H}=+241.82 \mathrm{~kJ} \end{aligned}
Q8
The equilibrium concentrations of the species in the reaction A+B⇌C+D\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D} are 2,3,102,3,10 and 6 mol6 \mathrm{~mol} L−1\mathrm{L}^{-1}, respectively at 300 K.ΔG0300 \mathrm{~K} . \Delta \mathrm{G}^{0} for the reaction is (R=2cal/mol K)(\mathrm{R}=2 \mathrm{cal} / \mathrm{mol} ~\mathrm{K})
A −137.26 cal-137.26 ~\mathrm{cal}
B −1381.80 cal-1381.80 ~\mathrm{cal}
C −13.73 cal-13.73 ~\mathrm{cal}
D 1372.60 cal1372.60 ~\mathrm{cal}
Correct Answer
Option B
Solution
A+B⇌C+D\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}
[A]=2 mol L−1[B]=3 mol L−1[C]=10 mol L−1[D]=6 mol L−1ΔG0=−2.303RTlog⁡Keq=−2.303RTlog⁡[C][D][A][B]=−2.303×2×300×log⁡10×62×3=−2.303×2×300×log⁡10=−1381.8 cal\begin{aligned} {[A] } & =2 \mathrm{~mol} \mathrm{~L}^{-1} \\ {[B] } & =3 \mathrm{~mol} \mathrm{~L}^{-1} \\ {[C] } & =10 \mathrm{~mol} \mathrm{~L}^{-1} \\ {[D] } & =6 \mathrm{~mol} \mathrm{~L}^{-1} \\ \Delta \mathrm{G}^{0} & =-2.303 \mathrm{RT} \log \mathrm{K}_{\mathrm{eq}} \\ & =-2.303 \mathrm{RT} \log \frac{[\mathrm{C}][\mathrm{D}]}{[\mathrm{A}][\mathrm{B}]} \\ & =-2.303 \times 2 \times 300 \times \log \frac{10 \times 6}{2 \times 3} \\ & =-2.303 \times 2 \times 300 \times \log 10 \\ & =-1381.8 \mathrm{~cal} \end{aligned}
Q9
Which amongst the following options is the correct relation between change in enthalpy and change in internal energy?
A ΔH=ΔU+ΔngRT\Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{n_g R T}
B ΔH−ΔU=−ΔnRT\Delta \mathrm{H}-\Delta \mathrm{U}=-\Delta \mathrm{nRT}
C ΔH+ΔU=ΔnR\Delta \mathrm{H}+\Delta \mathrm{U}=\Delta \mathrm{nR}
D ΔH=ΔU−ΔngRT\Delta \mathrm{H}=\Delta \mathrm{U}-\Delta \mathrm{n_g R T}
Correct Answer
Option A
Solution
ΔH=ΔU+ΔngRT\Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{n_g R T}
Q10
One mole of an ideal gas at 300 K is expanded isothermally from 1 L to 10 L volume. Δ\DeltaU for this process is : (Use R = 8.314 J k −-1 mol −-1 )
A 0 J
B 1260 J
C 2520 J
D 5040 J
Correct Answer
Option A
Solution
Δ\Delta

U = nC v

Δ\Delta

T For isothermal condition;

Δ\Delta

T = 0 ∴\therefore

Δ\Delta

U = 0

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