Practice Start Test

Thermodynamics

JEE Chemistry · 103 questions · Page 4 of 11 · Click an option or "Show Solution" to reveal answer

Q31
An ideal gas expands in volume from 1×\times10-3 m3 to 1 ×\times 10-2 m3 at 300 K against a constant pressure of 1×\times105 Nm-2. The work done is :
A -900 J
B 900 kJ
C 270 kJ
D -900 kJ
Correct Answer
Option A
Solution
w=−PΔV=−10−5(1×10−2−1×10−3)w = - P\Delta V = - {10^{ - 5}}\left( {1 \times {{10}^{ - 2}} - 1 \times {{10}^{ - 3}}} \right)
=−900J= - 900J
Q32
The enthalpies of combustion of carbon and carbon monoxide are -393.5 and -283 kJ mol-1 respectively. The enthalpy of formation of carbon monoxide per mole is :
A 110.5 kJ
B -110.5 kJ
C -676.5 kJ
D 676.5 kJ
Correct Answer
Option B
Solution
(i)   C+O2⇌CO2,\left( i \right)\,\,\,C + {O_2}\rightleftharpoons C{O_2},
  ΔH=−393.5 kJmol−1\,\,\Delta H = - 393.5\,kJmo{l^{ - 1}}
(ii)   CO+12O2⇌ CO2,\left( {ii} \right)\,\,\,CO + {1 \over 2}{O_2}\rightleftharpoons\,C{O_2},
        ΔH=−283.0 kJmol−1\,\,\,\,\,\,\,\,\Delta H = - 283.0\,kJmo{l^{ - 1}}

Operating

(i)−(ii),(i)-(ii),

we have

C+12O2→COC + {1 \over 2}{O_2} \to CO
        ΔH=−110.5 kJmol−1\,\,\,\,\,\,\,\,\Delta H = - 110.5\,kJmo{l^{ - 1}}
Q33
Consider an endothermic reaction, X →\to Y with the activation energies Eb and Ef for the backward and forward reactions, respectively. In general :
A Eb < Ef
B Eb > Ef
C Eb = Ef
D There is no definite relation between Eb and Ef
Correct Answer
Option A
Solution

Enthalpy of reaction

(ΔH)=Ea(f)−Ea(b)\left( {\Delta H} \right) = {E_{{a_{\left( f \right)}}}} - {E_{{a_{\left( b \right)}}}}

for an endothermic reaction

ΔH=+ve\Delta H = + ve

Hence for

ΔH\Delta H

to be negative

                     \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
                     \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
Ea(b)<Ea(f){E_{{a_{\left( b \right)}}}} < {E_{{a_{\left( f \right)}}}}
Q34
Consider the reaction: N2 + 3H2 →\to 2NH3 carried out at constant temperature and pressure. If ΔH\Delta H and ΔU\Delta U are the enthalpy and internal energy changes for the reaction, which of the following expressions is true?
A ΔH\Delta H > ΔU\Delta U
B ΔH\Delta H < ΔU\Delta U
C ΔH\Delta H = ΔU\Delta U
D ΔH\Delta H = 0
Correct Answer
Option B
Solution
ΔH=ΔU+ΔnRT\Delta H = \Delta U + \Delta nRT

for

   N2+3H2→2NH3\,\,\,{N_2} + 3{H_2} \to 2N{H_3}
   Δng=2−4=−2\,\,\,\Delta {n_g} = 2 - 4 = - 2

∴\therefore

   ΔH=ΔU−2RT   \,\,\,\Delta H = \Delta U - 2RT\,\,\,

or

   ΔU=ΔH+2RT\,\,\,\Delta U = \Delta H + 2RT

∴\therefore

   ΔU>ΔH\,\,\,\Delta U > \Delta H
Q35
If the bond dissociation energies of XY, X2 and Y2 (all diatomic molecules) are in the ratio of 1:1:0.5 and ΔHf\Delta H_f for the formation of XY is -200 kJ mole-1. The bond dissociation energy of X2 will be :
A 100 kJ mol-1
B 200 kJ mol-1
C 300 kJ mol-1
D 800 kJ mol-1
Correct Answer
Option D
Solution
X2+Y2→2XY,{X_2} + {Y_2} \to 2XY,
ΔH=2(−200).\Delta H = 2\left( { - 200} \right).

Let

xx

be the bond dissociation energy of

X2.{X_2}.

Then

ΔH=−400\Delta H = - 400
=ξx−x+ξy−y−2ξx−y= {\xi _{x - x}} + {\xi _{y - y}} - 2{\xi _{x - y}}
=x+0.5x−2x= x + 0.5x - 2x
=−0.5x= - 0.5x

or

   x=4000.5=800 kJ mol−1\,\,\,x = {{400} \over {0.5}} = 800\,kJ\,mo{l^{ - 1}}
Q36
An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If Ti is the initial temperature and Tf is the final temperature, which of the following statements is correct?
A (Tf)irrev > (Tf)rev
B (Tf)rev = (Tf)irrev
C Tf > Ti for reversible process but Tf = Ti for irreversible process
D Tf = Ti for both reversible and irreversible processes
Correct Answer
Option A
Solution

NOTE : In a reversible process the work done is greater than in irreversible process.

Hence the heat absorbed in reversible process would be greater than in the latter case.

So,

           \,\,\,\,\,\,\,\,\,\,\,
Tf(rev.)<Tf(irr.){T_f}\left( {rev.} \right) < {T_f}\left( {irr.} \right)
Q37
The process with negative entropy change is :
A Dissociation of CaSO4 (s) to CaO(s) and SO3(g)
B Dissolution of iodine in water
C Synthesis of ammonia from N2 and H2
D Sublimation of dry ice
Correct Answer
Option C
Solution

N2(g) + 3H2(g) ⇌\rightleftharpoons 2NH3(g);

Δ\Delta

ng < 0

Q38
The standard enthalpy of formation ΔfHo\Delta _fH^o at 298 K for methane, CH4(g), is –74.8 kJ mol–1. The additional information required to determine the average energy for C – H bond formation would be :
A the dissociation energy of H2 and enthalpy of sublimation of carbon
B latent heat of vapourization of methane
C the first four ionization energies of carbon and electron gain enthalpy of hydrogen
D the dissociation energy of hydrogen molecule, H2
Correct Answer
Option A
Solution

The standard enthalpy of formation of

CH4C{H_4}

is given by the equation :

                  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
                  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
C(s)+2H2(g)→CH4(g)C\left( s \right) + 2{H_2}\left( g \right) \to C{H_4}\left( g \right)

In order to calculate average energy for C – H bond formation we should know the following data.

C(graphite) →\to C(g) ;

ΔHfo\Delta H_f^o

= enthalpy of sublimation of carbon H2(g) →\to 2H(g) ;

Δ\Delta

H = bond dissociation energy of H2

Q39
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
Q40
In conversion of lime-stone to lime, CaCO3(s) →\to CaO(s) + CO2 (g) the vales of ∆H° and ∆S° are +179.1 kJ mol−1 and 160.2 J/K respectively at 298 K and 1 bar. Assuming that ∆H° do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is :
A 1008 K
B 1200
C 845 K
D 1118 K
Correct Answer
Option D
Solution
ΔG∘=ΔH∘−TΔS∘\Delta {G^ \circ } = \Delta {H^ \circ } - T\Delta {S^ \circ }

For a spontaneous reaction

ΔG∘<0\Delta {G^ \circ } < 0

or

   ΔH∘−TΔS∘<0\,\,\,\Delta {H^ \circ } - T\Delta {S^ \circ } < 0
⇒T>ΔH∘ΔS∘\Rightarrow T > {{\Delta {H^ \circ }} \over {\Delta {S^ \circ }}}
⇒T>179.3×103160.2>1117.9K≈1118K\Rightarrow T > {{179.3 \times {{10}^3}} \over {160.2}} > 1117.9K \approx 1118K
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