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Thermodynamics

NEET Chemistry · 93 questions · Page 9 of 10 · Click an option or "Show Solution" to reveal answer

Q81
The combustion of benzene(l) gives CO2(g) and H2O(l). Given that heat of combustion of benzene at constant volume is –3263.9 kJ mol–1 at 25oC; heat of combustion (in kJ mol–1) of benzene at constant pressure will be : (R = 8.314 JK–1 mol–1)
A –3267.6
B 4152.6
C –452.46
D 3260
Correct Answer
Option A
Solution

Formula of Heat of combination is

ΔH=Δu  +  Δng  RT\Delta H = \Delta u\,\, + \,\,\Delta ng\,\,RT

Where,

ΔH\Delta H

== Heat of combination at constant pressure

Δu =\Delta u\, =

Heat at constant volume

Δng\Delta {n_g}

change in number of moles for gaseous molecule. R = gas constant T = Temperature. The Required reaction C6H6(l) +

152{{15} \over 2}

O2(g) →\to 6CO2(g) + 3H2O(l) Here O2 and CO2 are gaseous molecules so to calculate

Δng\Delta {n_g}

we only consider those.

∴   \therefore\,\,\,
Δng\Delta {n_g}

== 6 −-

152{{15} \over 2}

= −-

32{3 \over 2}

Given

Δ\Delta

u = −- 3263.9 kJ mol−-1 R = 8.314 JK−- mol−-1 R = 8.314 JK−-1 mol−-1 = 8.314 ×\times 10−-3 kJ K−-1 mol−-1 T = 25o C = 25 + 273 K = 298 K So,

ΔH\Delta H

= −- 3263.9 +

(−32)\left( { - {3 \over 2}} \right)

×\times 8.314 ×\times 10−-3 ×\times 298 = −- 3267.6 kJ mol−-1

Q82
500 J of energy is transferred as heat to 0.5 mol of Argon gas at 298 K and 1.00 atm. The final temperature and the change in internal energy respectively are: Given: R = 8.3 J K-1 mol-1
A 368 K and 500 J
B 348 K and 300 J
C 378 K and 300 J
D 378 K and 500 J
Correct Answer
Option B
Solution

Amount of energy transpired as hat, Q=500 JQ=500 \mathrm{~J} Number of moles of Argon, n=0.5 moln=0.5 \mathrm{~mol} Temperature (Initial), Ti=298 KT_i=298 \mathrm{~K} Pressure, P=1.00 atmP=1.00 \mathrm{~atm} Final temperature, TF=T_F= ?

Change in internal energy, ΔV=\Delta V= ?

R=8.3 J K−1 mol−1R=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}

For a constant pressure process, the heat added is related to the temperature changes as Q=nCpΔTQ=n C_p \Delta T ....... (1)

 Also, the change in internal energy formula is ΔU=nCVΔT−(2).... (2)\begin{aligned} &\text{ Also, the change in internal energy formula is }\\ &\Delta U=n C_V \Delta T \quad-(2)\quad\text{.... (2)} \end{aligned}
 For monoatomic gas (Argon), Cv=32R So, Cp=Cv+R(Cp−Cv=R)=32R+R=52R\begin{aligned} &\text{ For monoatomic gas (Argon), }\\ &\begin{aligned} C_v & =\frac{3}{2} R \\ \text{ So, } C_p & =C_v+R \quad\left(C_p-C_v=R\right) \\ & =\frac{3}{2} R+R=\frac{5}{2} R \end{aligned} \end{aligned}
 In equation (1) ⇒Q=nCpΔTQ=nCp(TF−Ti)500 J=0.5 mol×52R(TF−Ti)R=8.3JK−1 mol−1500 J=0.5 mol×52×8.3 JK−1 mol−1(TF−Ti)\begin{aligned} &\text{ In equation (1) } \Rightarrow\\ &\begin{aligned} & Q=n C_p \Delta T \\ & Q=n C_p\left(T_F-T_i\right) \\ & 500 \mathrm{~J}=0.5 \mathrm{~mol} \times \frac{5}{2} R\left(T_F-T_i\right) \quad R=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \\ & 500 \mathrm{~J}=0.5 \mathrm{~mol} \times \frac{5}{2} \times 8.3 \mathrm{~J} K^{-1} \mathrm{~mol} ^{-1}\left(T_F-T_i\right) \end{aligned} \end{aligned}

TF−Ti=500 J0.5mol×52×8.3 J K1 mol−1T_F-T_i=\dfrac{500 \mathrm{~J}}{0.5 \mathrm{mol} \times \dfrac{5}{2} \times 8.3 \text{ J K}^1 \mathrm{~mol}^{-1}}

=5000.5×52×8.3k=48.2kTF=48.2k+298k=346.2k≈348k\begin{aligned} & =\frac{500}{0.5 \times \frac{5}{2} \times 8.3} \mathrm{k} \\ & =48.2 \mathrm{k} \\ T_F & =48.2 \mathrm{k}+298 \mathrm{k} \\ & =346.2 \mathrm{k} \\ & \approx 348 \mathrm{k} \end{aligned}
 In equation (2) ⇒ΔU=nCVΔT=0.5 mol×32R×48.2 K=0.5 mol×32×8.3 J K−1mol−1×48.2 K=300.045 J≈300 J\begin{aligned} &\text{ In equation (2) } \Rightarrow\\ &\begin{aligned} \Delta U & =n C_V \Delta T \\ & =0.5 \mathrm{~mol} \times \frac{3}{2} R \times 48.2 \mathrm{~K} \\ & =0.5 \mathrm{~mol} \times \frac{3}{2} \times 8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{mol^{-1}} \times 48.2 \mathrm{~K} \\ & =300.045 \mathrm{~J} \approx 300 \mathrm{~J} \end{aligned} \end{aligned}

The final temperature is, 348 k change in internal energy is 300 J So, correct answer is Option 2) 348 K and 300 J

Q83
During which of the following processes, does entropy decrease? (A) Freezing of water to ice at 0∘^\circC (B) Freezing of water to ice at −-10∘^\circC (C) N2(g) + 3H2(g) → \to 2NH3(g) (D) Adsorption of CO(g) on lead surface. (E) Dissolution of NaCl in water Choose the correct answer from the options given below :
A (A), (C) and (E) only
B (B) and (C) only
C (A), (B), (C) and (D) only
D (A) and (E) only
Correct Answer
Option C
Solution

A, B →\to Freezing of water will decrease entropy as particles will move closer and forces of attraction will increase.

This leads to decrease in randomness.

So entropy decrease.

C →\to No. of molecules decreasing D →\to Adsorption will lead to decrease in randomness of gaseous particles.

E →\to NaCl(s) →\to Na+(aq) + Cl–(aq)

Δ\Delta

S > 0 So, (A, B, C, D) decreases entropy.

Q84
At 25∘^\circC and 1 atm pressure, the enthalpy of combustion of benzene (I) and acetylene (g) are −- 3268 kJ mol−-1 and −-1300 kJ mol−-1, respectively. The change in enthalpy for the reaction 3 C2H2(g) →\to C6H6 (I), is :
A +324 kJ mol−-1
B +632 kJ mol−-1
C −- 632 kJ mol−-1
D −- 732 kJ mol−-1
Correct Answer
Option C
Solution

I.

C6H6(ℓ)+152O2( g)→6CO2( g)+3H2O(g)\mathrm{C}_{6} \mathrm{H}_{6}(\ell)+\dfrac{15}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 6 \mathrm{CO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})

ΔH1=−3268 kJ/mol\Delta \mathrm{H}_{1}=-3268 \mathrm{~kJ} / \mathrm{mol}

II.

C2H2( g)+52O2( g)→2CO2( g)+H2O(g)\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g})+\dfrac{5}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})

ΔH2=−1300 kJ/mol\Delta \mathrm{H}_{2}=-1300 \mathrm{~kJ} / \mathrm{mol}

III.

3C2H2( g)→C6H6(ℓ)ΔH33 \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{6} \mathrm{H}_{6}(\ell) \quad \Delta \mathrm{H}_{3} Applying Hess's law of constant heat summation

ΔH3=3×ΔH2−ΔH1=3×(−1300)−(−3268)=−632 kJ/mol\begin{aligned} \Delta \mathrm{H}_{3} &=3 \times \Delta \mathrm{H}_{2}-\Delta \mathrm{H}_{1} \\\\ &=3 \times(-1300)-(-3268) \\\\ &=-632 \mathrm{~kJ} / \mathrm{mol} \end{aligned}
Q85
Which one of the following equations does not correctly represent the first law of thermodynamics for the given processes involving an ideal gas? (Assume non-expansion work is zero)
A Adiabatic process : Δ\Delta U= – w
B Cyclic process : q = –w
C Isochoric process : Δ\Delta U= q
D Isothermal process : q = – w
Correct Answer
Option A
Solution

From 1st law of thermodynamics we know,

Δ\Delta

U = q + W Option A : In adiabatic process exchage of heat = 0 ∴\therefore q = 0 ∴\therefore From 1st law of thermodynaics,

Δ\Delta

U = W So option A is wrong.

Option B : U is a state function.

In cyclic process, initial state and final state both are same.

So change in all the state function in cyclic process will be zero.

∴\therefore

Δ\Delta

U = 0 ∴\therefore From 1st law of thermodynaics, q + W = 0 ⇒\Rightarrow q = -W So option B is correct..

Option C : In isochoric process volume (V) is constant.

So dV = 0.

We know, W =

−∫PexdV- \int {{P_{ex}}} dV

∴\therefore W = 0 ∴\therefore From 1st law of thermodynaics,

Δ\Delta

U = q So option C is correct. Option D : In isothermal process temerature (T) is constant. So dT = 0. We know,

Δ\Delta

U = nCvdT ∴\therefore

Δ\Delta

U = 0 ∴\therefore From 1st law of thermodynaics, q + W = 0 ⇒\Rightarrow q = -W So option D is correct.

Q86
One mole of an ideal gas expands isothermally and reversibly from 10dm310 \mathrm{dm}^3 to 20dm320 \mathrm{dm}^3 at 300 K . ΔU,q\Delta \mathrm{U}, \mathrm{q} and work done in the process respectively are Given: R=8.3 J K−1 mol−1\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} ln⁡10=2.3\ln 10=2.3 log⁡2=0.30\log 2=0.30 log⁡3=0.48\log 3=0.48
A 0,21.84 kJ,−1.726 J0,21.84 \mathrm{~kJ},-1.726 \mathrm{~J}
B 0,21.84 kJ,21.84 kJ0,21.84 \mathrm{~kJ}, 21.84 \mathrm{~kJ}
C 0,1.718 kJ,−1.718 kJ0,1.718 \mathrm{~kJ},-1.718 \mathrm{~kJ}
D 0,−17.18 kJ,1.718 J0,-17.18 \mathrm{~kJ}, 1.718 \mathrm{~J}
Correct Answer
Option C
Solution
(10 L,300 K)→n=1(20 L,300 K)−q=w=−nRTln⁡ V2 V1=−8.3×300×ln⁡(2010)=−1.718 kJ⇒q=1.718 kJw=−1.718 kJΔU=0(∵Δ T=0)\begin{aligned} & (10 \mathrm{~L}, 300 \mathrm{~K}) \xrightarrow{\mathrm{n}=1}(20 \mathrm{~L}, 300 \mathrm{~K}) \\ & -\mathrm{q}=\mathrm{w}=-\mathrm{nRT} \ln \frac{\mathrm{~V}_2}{\mathrm{~V}_1} \\ & =-8.3 \times 300 \times \ln \left(\frac{20}{10}\right) \\ & =-1.718 \mathrm{~kJ} \\ & \Rightarrow \mathrm{q}=1.718 \mathrm{~kJ} \\ & \mathrm{w}=-1.718 \mathrm{~kJ} \\ & \Delta \mathrm{U}=0(\because \Delta \mathrm{~T}=0) \end{aligned}
Q87
For silver, Cp(J K–1 mol–1) = 23 +0.01 T. If the temperature (T) of 3 moles of silver is raised from 300 K to 1000 K at 1 atm pressure, the value of ΔH\Delta H will be close to :
A 62 KJ
B 16 KJ
C 13 KJ
D 21 KJ
Correct Answer
Option A
Solution

Give that, n = 3 T1 = 300 T2 = 1000 Cp = 23 + 0.01T We know,

Δ\Delta

H =

∫T1T2nCpdT\int\limits_{{T_1}}^{{T_2}} {n{C_p}dT}

=

∫30010003(23+T100)dT\int\limits_{300}^{1000} {3\left( {23 + {T \over {100}}} \right)dT}

=

3[23T+T2200]30010003\left[ {23T + {{{T^2}} \over {200}}} \right]_{300}^{1000}

= 3[ 23(1000 - 300 +

1200{{1 \over {200}}}

((1000)2 - (300)2)] = 3[ 23 ×\times 700 +

700×1300200{{700 \times 1300} \over {200}}

] = 61950 J = 61.95 kJ

≃\simeq

62 kJ

Q88
The correct relationship between free energy change in a reaction and the corresponding equilibrium constant Kc is :
A - ΔG\Delta G = RT ln Kc
B ΔGo\Delta G^o = RT ln Kc
C - ΔGo\Delta G^o = RT ln Kc
D ΔG\Delta G = RT ln Kc
Correct Answer
Option C
Solution
ΔG∘=−RT 1nKc   \Delta {G^ \circ } = - RT\,1n{K_c}\,\,\,

or

   −ΔG∘=RT 1nKc\,\,\, - \Delta {G^ \circ } = RT\,1n{K_c}
Q89
The enthalpy changes for the following processes are listed below : Cl2(g) = 2Cl(g), 242.3 kJ mol–1 I2(g) = 2I(g), 151.0 kJ mol–1 ICl(g) = I(g) + Cl(g), 211.3 kJ mol–1 I2(s) = I2(g), 62.76 kJ mol–1 Given that the standard states for iodine and chlorine are I2(s) and Cl2(g), the standard enthalpy of formation for ICl(g) is :
A –14.6 kJ mol–1
B –16.8 kJ mol–1
C +16.8 kJ mol–1
D +244.8 kJ mol–1
Correct Answer
Option C
Solution
I2(s)+Cl2(g)→2ICl(g){{\rm{I}}_2}\left( s \right) + C{l_2}\left( g \right) \to 2{\rm{I}}Cl\left( g \right)
ΔA=[ΔI2(s)→l2(g)+ΔHI−l+ΔHCI−Cl]−\Delta A = \left[ {\Delta {{\rm{I}}_2}\left( s \right) \to {l_2}\left( g \right) + \Delta {H_{{\rm I} - l}} + \Delta {H_{C{\rm I} - Cl}}} \right] -
                           2[ΔHI−Cl]{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 2\left[ {\Delta {H_{{\rm{I}} - Cl}}} \right]
=151.0+242.3+62.76−2×211.3= 151.0 + 242.3 + 62.76 - 2 \times 211.3
=33.46= 33.46
ΔHf0(ICl)=33.462\Delta H_f^0\left( {{\rm{I}}Cl} \right) = {{33.46} \over 2}
=16.73  kJ/mol= 16.73\,\,kJ/mol
Q90
The true statement amongst the following is :
A S is a function of temperature but Δ\Delta S is not a function of temperature.
B Both S and Δ\Delta S are not functions of temperature.
C Both Δ\Delta S and S are functions of temperature.
D S is not a function of temperature but Δ\Delta S is a function of temperature.
Correct Answer
Option C
Solution
ΔS=∫dqrevT\Delta S = \int {{{d{q_{rev}}} \over T}}

S = Kln(w) Both entropy and change in entropy are function of temperature.

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