Question

For the reaction at 298 K,

2A + B→C

ΔH = 200 kJ mol–1 and ΔS = 0.2 kJ K–1 mol–1

At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range?

Answer 1

From the expression, ∆G = ∆H – T∆S Assuming the reaction at equilibrium, ∆T for the reaction would be: (∆G = 0 at equilibrium) T = 2000 K For the reaction to be spontaneous, ∆G must be negative. Hence, for the given reaction to be spontaneous, T should be greater than 2000 K.

Answer 2

Nice chapter dear ....

hey here is your solution .......

from the expression

∆G = ∆H - T∆S

Assuming the reaction at equilibrium, ∆T for the reaction would be:

T = ∆H -∆G 1/∆S

T = ∆H/∆S

T = 200/0.2

T = 1000

(∆G = 0 at equilibrium)

For the reaction to be spontaneous, ∆G must be negative. Hence, for the given reaction to be spontaneous, T should be greater than 1000 K.

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