Question

For the equilibrium, $A\left(g\right)\rightleftharpoons B\left(g\right),\Delta His-40$ kJ/mol. If the ratio of the activation energies of the forward $\left(E_f\right)$ and reverse $\left(E_b\right)$ reactions is $\frac{2}{3}$ then:

• A. $E_f=30$ kJ/mol, $E_b=70$ kJ/mol
• B. $E_f=70$ kJ/mol, $E_b=30$ kJ/mol
• C. $E_f=80$ kJ/mol, $E_b=120$ kJ/mol
• D. $E_f=60$ kJ/mol, $E_b=100$ kJ/mol

Answer 1

Answer: (C)

$A\left(g\right)\rightleftharpoons B\left(g\right);\Delta H=-40$ kJ/mole

$\Delta H=E_{a\left(f\right)}-E_{a\left(b\right)}=-40kJ$

$\frac{E_{a\left(f\right)}}{E_{a\left(b\right)}}=\frac{2}{3}$

$\Rightarrow E_{a\left(f\right)}=\frac{2}{3}{E}_{a\left(b\right)}$

$\Rightarrow \frac{2}{3}{E}_{a\left(b\right)}-E_{a\left(b\right)}=-\frac{E_{a\left(b\right)}}{3}=-40kJ$

$E_{a\left(b\right)}=+120kJ$

$E_{a\left(f\right)}=\frac{2}{3}\times 120=80kJ$