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Question: At 300 K, on addition of catalyst to a reaction, rate of reaction increases to e$^{20}$ times of its...

At 300 K, on addition of catalyst to a reaction, rate of reaction increases to e20^{20} times of its initial value, then difference of activation energy in absence of catalyst and in presence of catalyst is 'q' kcal/mole. The value of q is

(Take R = 2 cal mol−1^{-1} K−1^{-1})

Answer

12

Explanation

Solution

The rate constant (kk) of a reaction is related to the activation energy (EaE_a) and temperature (TT) by the Arrhenius equation:

k=Ae−Ea/RTk = A e^{-E_a / RT}

where AA is the pre-exponential factor and RR is the gas constant.

A catalyst increases the rate of reaction by lowering the activation energy. Let k1k_1 and Ea1E_{a1} be the rate constant and activation energy in the absence of a catalyst, respectively. Let k2k_2 and Ea2E_{a2} be the rate constant and activation energy in the presence of a catalyst, respectively. The pre-exponential factor AA is assumed to be the same in both cases.

In the absence of a catalyst:

k1=Ae−Ea1/RTk_1 = A e^{-E_{a1} / RT}

In the presence of a catalyst:

k2=Ae−Ea2/RTk_2 = A e^{-E_{a2} / RT}

We are given that the rate of reaction (and thus the rate constant) increases to e20e^{20} times its initial value upon addition of a catalyst.

So, k2k1=e20\frac{k_2}{k_1} = e^{20}

Divide the equation for k2k_2 by the equation for k1k_1:

k2k1=Ae−Ea2/RTAe−Ea1/RT\frac{k_2}{k_1} = \frac{A e^{-E_{a2} / RT}}{A e^{-E_{a1} / RT}}

e20=e(−Ea2RT−−Ea1RT)e^{20} = e^{\left(\frac{-E_{a2}}{RT} - \frac{-E_{a1}}{RT}\right)}

e20=e(Ea1−Ea2RT)e^{20} = e^{\left(\frac{E_{a1} - E_{a2}}{RT}\right)}

Taking the natural logarithm on both sides:

20=Ea1−Ea2RT20 = \frac{E_{a1} - E_{a2}}{RT}

The difference of activation energy in the absence of catalyst and in the presence of catalyst is q=Ea1−Ea2q = E_{a1} - E_{a2}.

So, 20=qRT20 = \frac{q}{RT}

q=20RTq = 20 RT

Given values:

Temperature, T=300 KT = 300 \text{ K}

Gas constant, R=2 cal mol−1 K−1R = 2 \text{ cal mol}^{-1} \text{ K}^{-1}

Substitute the values into the equation for qq:

q=20×(2 cal mol−1 K−1)×(300 K)q = 20 \times (2 \text{ cal mol}^{-1} \text{ K}^{-1}) \times (300 \text{ K})

q=12000 cal mol−1q = 12000 \text{ cal mol}^{-1}

The question asks for the value of qq in kcal/mole.

Since 1 kcal=1000 cal1 \text{ kcal} = 1000 \text{ cal}:

q=120001000 kcal mol−1q = \frac{12000}{1000} \text{ kcal mol}^{-1}

q=12 kcal mol−1q = 12 \text{ kcal mol}^{-1}