Question

For a reaction A (g) $\rightleftharpoons$B(g) at equilibrium. The partial pressure of B is found to be one fourth of the partial pressure of A. The value of $\Delta$G⁰ of the reaction A $\rightarrow$B is

• A. RT $\mathcal{l}$n 4
• B. – RT $\mathcal{l}$n 4
• C. RT log 4
• D. – RT log 4

Answer 1

Answer: (A)

RT $\mathcal{l}$n 4

Answer 2

A (g) $\rightleftharpoons$ B (g) PB = $\frac{1}{4}$PA PA = 4PB

Kp = $\frac{P_{B}}{P_{A}}$ = $\frac{P_{A}/4}{P_{A}}$ = $\frac{1}{4}$

At equilibrium, $\Delta$G = 0.