Consider the reaction textA arrow 2 text B + textC, Δ textH = - 1 5 text kcal . The energy of activation of backward reaction is 20 kcal mol-1. In presence of catalyst, the energy of activation of forward reaction is 3 kcal mol-1. At 400 K the catalyst causes the rate of the forward reaction to increase by the number of times equal to-

Question

Consider the reaction textA arrow 2 text B + textC, Δ textH = - 1 5 text kcal . The energy of activation of backward reaction is 20 kcal mol-1. In presence of catalyst, the energy of activation of forward reaction is 3 kcal mol-1. At 400 K the catalyst causes the rate of the forward reaction to increase by the number of times equal to- (A) e3.5 (B) e2.5 (C) e-2.5 (D) e2.303

Tardigrade Admin · Accepted Answer

mathrmE mathrma( mathrmf)- mathrmE mathrma( mathrmb)=Δ mathrmH=-15 mathrmkcal ⇒ Ea(f)=-15+20=5 mathrmkcal ( mathrmk text (catalyst) / mathrmk mathrma)= mathrme mathrmE mathrma- mathrmE text a(catalyst) / mathrmRT = mathrme(5-3) × 103 / 2 × 400 = texte2 text. 5

e^3.5

e^2.5

e^-2.5

e^2.303

$E_{a (f)} - E_{a (b)} = Δ H = - 15 kcal$

$\Rightarrow E_{a (f)} = - 15 + 20 = 5 kcal$

$\frac{k _{(catalyst)}}{k _{a}} = e^{E_{a} - E_{a(catalyst)} / RT}$

$= e^{(5 - 3) \times 1 0^{3} /2 \times 400}$

$= e^{2.5}$

e3.5

e2.5

e-2.5

e2.303

Ea(f)​−Ea(b)​=ΔH=−15kcal ⇒Ea(f)​=−15+20=5kcal ka​k(catalyst) ​​=eEa​−Ea(catalyst) ​/RT =e(5−3)×103/2×400 =e2.5

e^3.5

e^2.5

e^-2.5

e^2.303

$E_{a (f)} - E_{a (b)} = Δ H = - 15 kcal$

$\Rightarrow E_{a (f)} = - 15 + 20 = 5 kcal$

$\frac{k _{(catalyst)}}{k _{a}} = e^{E_{a} - E_{a(catalyst)} / RT}$

$= e^{(5 - 3) \times 1 0^{3} /2 \times 400}$

$= e^{2.5}$