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Thermodynamics question

2012 · Shift 0 · Q7
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Thermodynamics question

2012 · Shift 0 · Q7

JEE MainChemistryThermodynamicsMCQ+4 / −1
The incorrect expression among the following is :
  1. A
    ΔGsystemΔStotal=−T{{\Delta {G_{system}}} \over {\Delta {S_{total}}}} = - TΔStotal​ΔGsystem​​=−T
  2. B
    In isothermal process wreversible{w_{reversible}}wreversible​=−nRT ln⁡ VfVi- nRT\,\ln \,{{{V_f}} \over {{V_i}}}−nRTlnVi​Vf​​
  3. C
    In K =ΔHo−TΔSoRTK\, = {{\Delta {H^o} - T\Delta {S^o}} \over {RT}}K=RTΔHo−TΔSo​
  4. D
    K = e−ΔGo/RTK\, = \,{e^{ - \Delta {G^o}/RT}}K=e−ΔGo/RT
View written solutionFree

Correct answer: C

  1. Check option A

For a process at constant temperature and pressure,

ΔGsystem=−T ΔStotal\Delta G_{\text{system}} = -T\,\Delta S_{\text{total}}ΔGsystem​=−TΔStotal​

So,

ΔGsystemΔStotal=−T\frac{\Delta G_{\text{system}}}{\Delta S_{\text{total}}} = -TΔStotal​ΔGsystem​​=−T

Hence, A is correct.

  1. Check option B

For a reversible isothermal expansion/compression of an ideal gas,

wrev=−∫ViVfP dVw_{\text{rev}} = -\int_{V_i}^{V_f} P\,dVwrev​=−∫Vi​Vf​​PdV

Using ideal gas equation for isothermal process,

P=nRTVP = \frac{nRT}{V}P=VnRT​

Therefore,

wrev=−∫ViVfnRTV dV=−nRTln⁡(VfVi)w_{\text{rev}} = -\int_{V_i}^{V_f} \frac{nRT}{V}\,dV = -nRT\ln\left(\frac{V_f}{V_i}\right)wrev​=−∫Vi​Vf​​VnRT​dV=−nRTln(Vi​Vf​​)

Hence, B is correct.

  1. Check option C

We know,

ΔGo=ΔHo−TΔSo\Delta G^o = \Delta H^o - T\Delta S^oΔGo=ΔHo−TΔSo

Also, the relation between equilibrium constant and standard Gibbs free energy is

ΔGo=−RTln⁡K\Delta G^o = -RT\ln KΔGo=−RTlnK

So,

ln⁡K=−ΔGoRT=−ΔHo−TΔSoRT\ln K = -\frac{\Delta G^o}{RT} = -\frac{\Delta H^o - T\Delta S^o}{RT}lnK=−RTΔGo​=−RTΔHo−TΔSo​

Thus,

K=e−ΔGoRT=e−ΔHo−TΔSoRTK = e^{-\frac{\Delta G^o}{RT}} = e^{-\frac{\Delta H^o - T\Delta S^o}{RT}}K=e−RTΔGo​=e−RTΔHo−TΔSo​

But option C states

K=ΔHo−TΔSoRTK = \frac{\Delta H^o - T\Delta S^o}{RT}K=RTΔHo−TΔSo​

This is incorrect because the correct relation is for ln⁡K\ln KlnK, not directly for KKK.

Hence, C is incorrect.

  1. Check option D

From

ΔGo=−RTln⁡K\Delta G^o = -RT\ln KΔGo=−RTlnK

we get

K=e−ΔGo/RTK = e^{-\Delta G^o/RT}K=e−ΔGo/RT

Hence, D is correct.

  1. Final conclusion

The incorrect expression is:

C\boxed{\text{C}}C​

This matches the stored correct answer.

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