For a reaction A(g) leftharpoons B(g) at equilibrium, the partial pressure of B is found to be one fourth of the partial pressure of A. The value of Δ G° for the reaction A arrow B is

Question

For a reaction A(g) leftharpoons B(g) at equilibrium, the partial pressure of B is found to be one fourth of the partial pressure of A. The value of Δ G° for the reaction A arrow B is (A) R T ln 4 (B) -R T ln 4 (C) R T log 4 (D) -R T log 4

Tardigrade Admin · Accepted Answer

Equilibrium constant, K=(pB/pA)=(1/4) Δ G°=-R T ln K=-R T ln (1/4)=R T ln 4

Q. For a reaction $A_{(g)} ⇌ B_{(g)}$ at equilibrium, the partial pressure of $B$ is found to be one fourth of the partial pressure of $A$ . The value of $Δ G^{\circ}$ for the reaction $A \to B$ is

$RT ln 4$

$- RT ln 4$

$RT lo g 4$

$- RT lo g 4$

Equilibrium constant, $K = \frac{p _{B}}{p _{A}} = \frac{1}{4}$

$Δ G^{\circ} = - RT ln K = - RT ln \frac{1}{4} = RT ln 4$

Q. For a reaction A(g)​⇌B(g)​ at equilibrium, the partial pressure of B is found to be one fourth of the partial pressure of A. The value of ΔG∘ for the reaction A→B is

RTln4

−RTln4

RTlog4

−RTlog4