In a one-litre flask, 6 moles of A undergoes the reaction A (   g ) ⇌ P (   g ) . The progress of product formation at two temperatures (in Kelvin), T 1 and T 2 , is shown in the figure: If T 1 = 2 T 2 and ΔG 2 o - ΔG 1 o = RT 2 lnx , then the value of x is [ ΔG 1 o and ΔG 2 o are standard Gibb's free energy change for the reaction at temperatures T 1 and T 2 , respectively.]
A. A
The equilibrium moles can be calculated as follows,                                                           A g ⇌ P g initial                                             6                           0 At   equilibrium               6 - y                 y at T 1    y = 4    at T 2    y = 2 ∴ K eq 1 = 4 2 = 2                                      ∴ K eq 2 = 2 4 = 1 2 The relation between Gibbs free energy and equilibrium constant, ∆ G o = - nRTlnK eq ΔG 1 o = - RT 1 ln K eq 1 ΔG 1 o = - 2 RT 2 ln K eq 1 [Given: T 1 = 2 T 2 ] ΔG 2 o = - RT 2 ln K eq 2 ∴ ΔG 2 o - ΔG 1 o = RT 2 ln K eq 1 2 K eq 2 = RT 2 ln 2 2 1 2 = RT 2 ln 8 ∴ ΔG 2 o - ΔG 1 o = RTlnx has x = 8
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