Discussion Forum

Consider the following reversible reaction,$A(g)+B(g)\rightleftharpoons AB (g)$. The activation energy of the backward reaction exceeds that of the forward reaction by 2 RT (in J mol^-1). If the pre-exponential factor of the forward  reaction is times that of the reverse reaction, the absolute value of $\triangle G^{\ominus}$  ( in J mol^-1) for the reaction at 300K is.................                     (Given; ln(2)=0.7 RT =2500J mol^-1 at 300K and G is the Gibbs energy )

Answer:

Explanation:

 For the reaction,

 $A(g)+B(g)\rightleftharpoons AB (g)$

 Given $E_{a_{b}}=E_{a_{f}}+2RT$

$E_{a_{b}}-E_{a_{f}}=2RT$

 Further, $A_{f}=4A_{b}$      or    $\frac{A_{f}}{A_{b}}=4$

   Now, rate constant for forward reaction,

  $k_{f}=A_{f}e^{\frac{-E_{af}}{RT}}$

 Like wise, rate constant for backward reaction

 $k_{b}=A_{b}e^{\frac{-E_{ab}}{RT}}$

At equillibrium,  Rate of forward reaction = Rate of backward reaction

 ie., $k_{f}=k_{b}$

or   $\frac{k_{f}}{k_{b}}= k_{eq}$

 So,  $k_{eq}= \frac{A_{f}e^{-E_{\frac{a_{f}}{RT}}}}{A_{b}e^{-E_{\frac{a_{b}}{RT}}}}$

  =   $\frac{A_{f}}{A_{b}}e^{\frac{-(E_{a_{f}}-E_{a_{b})}}{RT}}$

 After putting the given values,  $k_{eq_{}}=4e^{2}$

 (as   $E_{a_{b}}-E_{a_{f}}=2RT$  and       $\frac{A_{f}}{A_{b}}=4$ )

  Now,

Δ G⁰ = -RT ln K_eq

= -2500 ln(4 e²)

= -2500 (ln 4 +ln e² )

= -2500 (1.4 +2)

= -2500 × 3.4

= -8500 J/mol

Absolute value = 8500 J/mol