If the rate constants of a reaction at 500 K | vedclass

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

Question

(A) The Arrhenius equation is given by $\log \left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 \ R} \left(\frac{T_2 - T_1}{T_1 \ T_2}\right)$.Given: $k_

Vedclass · Accepted Answer

$49.49 \ kJ \ mol^{-1}$

Vedclass · Answer

(A) The Arrhenius equation is given by $\log \left(\frac{k_2}{k_1}ight) = \frac{E_a}{2.303 \ R} \left(\frac{T_2 - T_1}{T_1 \ T_2}ight)$.Given: $k_1 = 0.002 \ s^{-1}$,$T_1 = 500 \ K$,$k_2 = 0.06 \ s^{-1}$,$T_2 = 700 \ K$,$R = 8.314 \ J \ mol^{-1} \ K^{-1}$.Substituting the values:$\log \left(\frac{0.06}{0.002}ight) = \frac{E_a}{2.303 	imes 8.314} \left(\frac{700 - 500}{700 	imes 500}ight)$$\log(30) = \frac{E_a}{19.147} \left(\frac{200}{350000}ight)$$1.477 = \frac{E_a}{19.147} 	imes \frac{2}{3500}$$E_a = \frac{1.477 	imes 19.147 	imes 3500}{200} \approx 49.49 \ kJ \ mol^{-1}$.

If the rate constants of a reaction at $500 \ K$ and $700 \ K$ are $0.002 \ s^{-1}$ and $0.06 \ s^{-1}$,respectively,the value of activation energy is $(R=8.314 \ J \ mol^{-1} \ K^{-1}, \log 3=0.477)$.

Similar Questions

In a reaction,for every $10 ^\circ C$ rise in temperature,the rate is doubled. If the temperature is increased from $10 ^\circ C$ to $100 ^\circ C$,the rate of the reaction will become ......... times.

Rate constant of a reaction is given as $\log_{10} K = -\frac{2000}{T} + 6.0$. Activation energy is ....... $kcal$.

For the forward reaction $X \rightarrow Y$,the activation energy is $60 \ kJ \ mol^{-1}$ and $\Delta H = -20 \ kJ \ mol^{-1}$. What is the activation energy for the reverse reaction $Y \rightarrow X$ in $kJ \ mol^{-1}$?

Which of the following statements is not true according to the collision theory of reaction rates?

In the given graph,$E_{a}$ for the reverse reaction will be (in $kJ$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module