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Thermodynamics question

2021 · 26 Feb · Shift 1 · Q16
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Thermodynamics question

2021 · 26 Feb · Shift 1 · Q16

JEE MainChemistryThermodynamicsNumerical+4 / −1
For a chemical reaction A + B ⇌ C + D (ΔrHΘ{\Delta _r}{H^\Theta }Δr​HΘ = 80 kJ mol −-− 1) the entropy change ΔrSΘ{\Delta _r}{S^\Theta }Δr​SΘ depends on the temperature T (in K) as ΔrSΘ{\Delta _r}{S^\Theta }Δr​SΘ = 2T (J K −-− 1mol −-− 1). Minimum temperature at which it will become spontaneous is ‾\underline{\hspace{2cm}}​ K. (Integer)
Numerical answer
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Correct answer: 200

  1. For spontaneity at constant temperature and pressure, the criterion is

ΔrGΘ=ΔrHΘ−TΔrSΘ<0\Delta_r G^\Theta = \Delta_r H^\Theta - T\Delta_r S^\Theta < 0Δr​GΘ=Δr​HΘ−TΔr​SΘ<0

The minimum temperature at which the reaction becomes spontaneous is obtained at the boundary:

ΔrGΘ=0\Delta_r G^\Theta = 0Δr​GΘ=0

  1. Given:

ΔrHΘ=80 kJ mol−1=80000 J mol−1\Delta_r H^\Theta = 80\ \text{kJ mol}^{-1} = 80000\ \text{J mol}^{-1}Δr​HΘ=80 kJ mol−1=80000 J mol−1

and

ΔrSΘ=2T J K−1mol−1\Delta_r S^\Theta = 2T\ \text{J K}^{-1}\text{mol}^{-1}Δr​SΘ=2T J K−1mol−1

  1. Substitute into the Gibbs free energy expression:

ΔrGΘ=80000−T(2T)\Delta_r G^\Theta = 80000 - T(2T)Δr​GΘ=80000−T(2T)

ΔrGΘ=80000−2T2\Delta_r G^\Theta = 80000 - 2T^2Δr​GΘ=80000−2T2

For the minimum temperature, set this equal to zero:

80000−2T2=080000 - 2T^2 = 080000−2T2=0

2T2=800002T^2 = 800002T2=80000

T2=40000T^2 = 40000T2=40000

T=200 KT = 200\ \text{K}T=200 K

(Temperature must be positive.)

  1. Therefore, the reaction becomes spontaneous for

T>200 KT > 200\ \text{K}T>200 K

So the minimum temperature is

200 K\boxed{200\ \text{K}}200 K​

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