Question

Consider the reaction A ⇌ B at 1000 K. At time ’, the temperature of the system was increased to 2000 K and the system was allowed to reach equilibrium. Throughout this experiment, the partial pressure of A was maintained at 1 bar. Given below is the plot of the partial pressure of B with time.
What is the ratio of the standard Gibbs energy of the reaction at 1000 K to that at 2000 K?

Answer 1

we have to find the ratio of the standard Gibbs energy of the reaction at 1000K to that at 2000K.

solution : consider the reaction, A ⇔B

from graph, equilibrium constant at 1000K,

K₁ = 10

and equilibrium constant at 2000K,

k₂ = 100

now using formula, ∆G = -RTlnK

so, Gibbs energy of reaction at 1000K , ∆G₁ = -R(1000)ln(10)

Gibbs energy of reaction at 2000K, ∆G₂ = -R(2000)ln(100)

now ∆G₁/∆G₂ = [-R(1000)ln(10)]/[-R(2000)ln(100)]

= (1/4)

= 0.25

Therefore the ratio of the standard Gibbs energy for the reaction at 1000K to that at 2000K is 0.25

Answer 2

Answer:

ratio = 1/4

Explanation:

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