Question

For following reactions
$A\stackrel{700K}{⟶}$ Product;
$A\stackrel{catalyst500K}{⟶}$Product
It was found that the $E_a$ is decreased by 30 kJ/mol in the presence of catalyst. If the rate remains unchanged , the activation energy for catalysed reaction is (Assume pre-exponenetial factor is same)

• A. 75 kJ/mol
• B. 135 kJ/mol
• C. 105 kJ/mol
• D. 198 kJ/mol

Answer 1

The correct option is A 75 kJ/mol

$K=Ae(-\frac{E_a}{RT})$
$K_{\text{catalyst}}=K_{\text{without catalyst}}$
$Ae(-\frac{\left(E_a\right)c}{R{T}_{500K}})=Ae(-\frac{\left(E_A\right)}{R{T}_{700K}})-\frac{\left(E_a\right)c}{R{T}_{500K}}=-\frac{\left(E_a\right)}{R{T}_{700K}}\left(E_a\right)c=E_a-30-\frac{(E_a-300)}{T_{500K}}=-\frac{\left(E_a\right)}{T_{700K}}$
On solving $E_a$ = $105kJmo{l}^{-1}$
Activation energy of the catalysed reaction = 105 - 30 = 75 $kJ/mole$