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For the Reaction at 298 K,2a + B → C at What Temperature Will the Reaction Become Spontaneous Considering δH And δS To Be Constant Over the Temperature Range? - CBSE (Science) Class 11 - Chemistry

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Question

For the reaction at 298 K,

2A + B → C

Δ= 400 kJ mol–1 and Δ= 0.2 kJ K–1 mol–1

At what temperature will the reaction become spontaneous considering Δand Δto be constant over the temperature range?

Solution 1

From the expression,

ΔG = ΔH – TΔS

Assuming the reaction at equilibrium, ΔT for the reaction would be:

`T = (triangle h - triangle G)1/(triangle S)`

= `(triangle H)/ (triangle S) (triangle G = 0 " at equilibrium")`

` = (400 kJmol^(-1))/(0.2 kJ K^(-1) mol^(-1))`

T = 2000 K

For the reaction to be spontaneous, ΔG must be negative. Hence, for the given reaction to be spontaneous, T should be greater than 2000 K.

Solution 2

As per the Gibbs Helmholtz equation:

ΔG = Δ H- TΔ S For ΔG=0 ; ΔH=TΔS or T=ΔH/ΔS

T = (400 KJ mol-1)/(0.2 KJ K-1 mol-1) = 2000 k

Thus, reaction will be in a state of equilibrium at 2000 K and will be spontaneous above this temperature.

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Solution For the Reaction at 298 K,2a + B → C at What Temperature Will the Reaction Become Spontaneous Considering δH And δS To Be Constant Over the Temperature Range? Concept: Spontaneity - Is Decrease in Enthalpy a Criterion for Spontaneity.
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