Question

For the system $A\left(g\right)\rightleftharpoons B\left(g\right),△H$for the forward reaction is $-33KJ/mol$ (Note: $△H=△E$ in this case). What will be the equilibrium constant $K$at 300 K. If the activation energies $E_f$ & $E_b$ are in the ratio 20:31, Calculate $E_f$ and $E_b$ at this temperature? (Assume that the pre-exponential factor is the same for the forward and backward reactions)

• A. $K=5.572\times {10}^5$
• B. $K=5.572\times {10}^4$
• C. $K=557.2\times {10}^4$
• D. None of these

Answer 1

The correct option is A $K=5.572\times {10}^5$

$E_f-E_b=\Delta H\to \left(1\right)$

Let $E_f$ be 20x & $E_b$ be 31x.

where x is the common factor.

20x-31x = -33

-11x = -33

x = 3

Now, $E_f$ = 60 $KJ/mol$ & $E_b$ = 93 $KJ/mol$

Using arrhenius equation.

$K_f={Ae}^{\frac{-60000}{\left(300\right)R}}$

$K_b={Ae}^{\frac{-93000}{300R}}$

By definition of equilibrium constant $K_{eq}=\frac{K_f}{K_b}$

$K_{eq}=e^{\frac{-60000}{300R}+\frac{93000}{300R}}$

$\Rightarrow e^{\frac{33000}{300R}}\Rightarrow e^{110/8.314}$

$⟶$ It is $5.572\times {10}^5$