The activation energy of one of the reactions | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Using the Arrhenius equation: $\ln \left(\frac{k_{310}}{k_{300}}\right) = \frac{E_a}{R} \left(\frac{1}{T_{300}} - \frac{1}{T_{310}}\right)$ $\ln \

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Using the Arrhenius equation: $\ln \left(\frac{k_{310}}{k_{300}}ight) = \frac{E_a}{R} \left(\frac{1}{T_{300}} - \frac{1}{T_{310}}ight)$ $\ln \left(\frac{k_{310}}{k_{300}}ight) = \frac{532611}{8.3} 	imes \left(\frac{310 - 300}{310 	imes 300}ight)$ $\ln \left(\frac{k_{310}}{k_{300}}ight) = \frac{532611}{8.3} 	imes \frac{10}{93000} = 64170 	imes \frac{10}{93000} \approx 6.9$ Since $\ln 10 = 2.3$,then $6.9 = 3 	imes 2.3 = 3 	imes \ln 10 = \ln 10^3$. Therefore,$\frac{k_{310}}{k_{300}} = 10^3$,which means $k_{300} = 10^{-3} \, k_{310}$. Comparing this with $k_{300} = x 	imes 10^{-3} \, k_{310}$,we get $x = 1$.

Similar Questions

What does $P \cdot Z_{AB} \cdot e^{-\frac{E_a}{RT}}$ indicate in the rate equation?

Explain the effect of an increase in temperature on the rate of reaction and the rate constant.

Which one of the following is true for an exothermic reaction $A \rightleftharpoons B$,if $E_f$ and $E_b$ are the activation energies of forward and backward reactions respectively?

The slope of the Arrhenius plot $(\ln k \, vs \, \frac{1}{T})$ of a first-order reaction is $-5 \times 10^{3} \, K$. The value of $E_{a}$ of the reaction is: (in $kJ \, mol^{-1}$)
$[\text{Given } R = 8.314 \, J \, K^{-1} \, mol^{-1}]$

For a reaction where the activation energies of the forward and reverse directions are equal in value,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module