Question

For the equilibrium, $A\left(g\right)\rightleftharpoons B\left(g\right), \, \Delta H is \, -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

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

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

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

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

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

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