Medium
For a first-order reaction $A \to P$,the rate constant equation is given by $\log K = -2000 \, (1/T) + 6.0$. The pre-exponential factor $A$ and the activation energy $E_a$ are,respectively:
- A$1.0 \times 10^{-6} \, s^{-1}$ and $92 \, kJ/mol$
- B$6 \, s^{-1}$ and $16.6 \, kJ/mol$
- C$1.0 \times 10^6 \, s^{-1}$ and $38.3 \, kJ/mol$
- D$1.0 \times 10^6 \, s^{-1}$ and $16.6 \, kJ/mol$
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Similar Questions
Two reactions $R_1$ and $R_2$ have identical pre-exponential factors. Activation energy of $R_1$ exceeds that of $R_2$ by $10 \, kJ \, mol^{-1}.$ If $k_1$ and $k_2$ are rate constants for reactions $R_1$ and $R_2$ respectively at $300 \, K,$ then $\ln (k_2/k_1)$ is equal to :
$(R=8.314 \, J \, mol^{-1} \, K^{-1})$
$(R=8.314 \, J \, mol^{-1} \, K^{-1})$
DifficultJEE MAIN 2017
View SolutionIn the Arrhenius equation,$k = A e^{-E_a/RT}$,the Arrhenius constant $A$ will be equal to the rate constant when
Medium
View SolutionThe rate constant for the decomposition of $N_{2}O_{5}$ at various temperatures is given below:
$T / ^{\circ}C$ $0$ $20$ $40$ $60$ $80$ $10^{5} \times k / s^{-1}$ $0.0787$ $1.70$ $25.7$ $178$ $2140$
Draw a graph between $\ln k$ and $1 / T$ and calculate the values of $A$ and $E_{a}.$ Predict the rate constant at $30^{\circ}C$ and $50^{\circ}C$.
| $T / ^{\circ}C$ | $0$ | $20$ | $40$ | $60$ | $80$ |
| $10^{5} \times k / s^{-1}$ | $0.0787$ | $1.70$ | $25.7$ | $178$ | $2140$ |
Draw a graph between $\ln k$ and $1 / T$ and calculate the values of $A$ and $E_{a}.$ Predict the rate constant at $30^{\circ}C$ and $50^{\circ}C$.
Difficult
View SolutionActivation energy $(E_a)$ and rate constants $(k_1)$ and $(k_2)$ of a chemical reaction at two different temperatures $(T_1)$ and $(T_2)$ are related by
MediumAIPMT 2012
View SolutionThe temperature at which the rate constants of the two gaseous reactions given below become equal is . . . . . . $K$. (Nearest integer).
$X \longrightarrow Y \quad k_1 = 10^6 e^{\frac{-30000}{T}}$
$P \longrightarrow Q \quad k_2 = 10^4 e^{\frac{-24000}{T}}$
Given: $\ln 10 = 2.303$
$X \longrightarrow Y \quad k_1 = 10^6 e^{\frac{-30000}{T}}$
$P \longrightarrow Q \quad k_2 = 10^4 e^{\frac{-24000}{T}}$
Given: $\ln 10 = 2.303$
DifficultJEE MAIN 2026
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