Difficult
The rate constant of a reaction is $2 \times 10^{-3} \ min^{-1}$ at $300 \ K$. When the temperature is increased by $20 \ K$,the rate constant becomes three times its initial value. Calculate the activation energy $(E_a)$ of the reaction. Also,determine the rate constant at $310 \ K$.
(N/A) Given: $k_1 = 2 \times 10^{-3} \ min^{-1}$ at $T_1 = 300 \ K$,$k_2 = 3 \times k_1 = 6 \times 10^{-3} \ min^{-1}$ at $T_2 = 320 \ K$.
Using the Arrhenius equation: $\ln(k_2/k_1) = (E_a/R) \times (1/T_1 - 1/T_2)$.
$\ln(3) = (E_a / 1.987) \times (1/300 - 1/320)$.
$1.0986 = (E_a / 1.987) \times (20 / 96000)$.
$E_a = 1.0986 \times 1.987 \times 4800 \approx 10480 \ cal/mol$.
To find $k$ at $310 \ K$: $\ln(k_3/k_1) = (E_a/R) \times (1/T_1 - 1/T_3)$.
$\ln(k_3 / 2 \times 10^{-3}) = (10480 / 1.987) \times (1/300 - 1/310) = 5274.3 \times (10 / 93000) = 0.5671$.
$k_3 / 2 \times 10^{-3} = e^{0.5671} = 1.763$.
$k_3 = 3.526 \times 10^{-3} \ min^{-1}$.
Using the Arrhenius equation: $\ln(k_2/k_1) = (E_a/R) \times (1/T_1 - 1/T_2)$.
$\ln(3) = (E_a / 1.987) \times (1/300 - 1/320)$.
$1.0986 = (E_a / 1.987) \times (20 / 96000)$.
$E_a = 1.0986 \times 1.987 \times 4800 \approx 10480 \ cal/mol$.
To find $k$ at $310 \ K$: $\ln(k_3/k_1) = (E_a/R) \times (1/T_1 - 1/T_3)$.
$\ln(k_3 / 2 \times 10^{-3}) = (10480 / 1.987) \times (1/300 - 1/310) = 5274.3 \times (10 / 93000) = 0.5671$.
$k_3 / 2 \times 10^{-3} = e^{0.5671} = 1.763$.
$k_3 = 3.526 \times 10^{-3} \ min^{-1}$.
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Similar Questions
$A$ reaction takes place in three steps with individual rate constant and activation energy as follows: $Step$ $Rate \ constant$ $Activation \ energy$ $Step-1$ $k_1$ $E_{a1} = 180 \ kJ/mol$ $Step-2$ $k_2$ $E_{a2} = 80 \ kJ/mol$ $Step-3$ $k_3$ $E_{a3} = 50 \ kJ/mol$
If overall rate constant,$k = (\frac{k_1 k_2}{k_3})^{2/3}$,then overall activation energy of the reaction will be .......... $kJ/mol$.
| $Step$ | $Rate \ constant$ | $Activation \ energy$ |
| $Step-1$ | $k_1$ | $E_{a1} = 180 \ kJ/mol$ |
| $Step-2$ | $k_2$ | $E_{a2} = 80 \ kJ/mol$ |
| $Step-3$ | $k_3$ | $E_{a3} = 50 \ kJ/mol$ |
Difficult
View Solution$A$ reaction takes place in three steps with individual rate constant and activation energy,
Step Rate constant and Activation energy $Step \ 1$ $k_1, E_{a_1} = 180 \ kJ \ mol^{-1}$ $Step \ 2$ $k_2, E_{a_2} = 80 \ kJ \ mol^{-1}$ $Step \ 3$ $k_3, E_{a_3} = 50 \ kJ \ mol^{-1}$
Overall rate constant,$k = (k_1 k_2 / k_3)^{2/3}$. The overall activation energy of the reaction will be ........ $kJ \ mol^{-1}$.
| Step | Rate constant and Activation energy |
| $Step \ 1$ | $k_1, E_{a_1} = 180 \ kJ \ mol^{-1}$ |
| $Step \ 2$ | $k_2, E_{a_2} = 80 \ kJ \ mol^{-1}$ |
| $Step \ 3$ | $k_3, E_{a_3} = 50 \ kJ \ mol^{-1}$ |
Overall rate constant,$k = (k_1 k_2 / k_3)^{2/3}$. The overall activation energy of the reaction will be ........ $kJ \ mol^{-1}$.
Difficult
View Solution$A$ catalyst
For a complex reaction $A \xrightarrow{K} \text{products}$,where $Ea_1 = 180 \ kJ/mol$,$Ea_2 = 80 \ kJ/mol$,and $Ea_3 = 50 \ kJ/mol$,the overall rate constant $K$ is related to individual rate constants by the equation $K = (\frac{K_1 \cdot K_2}{K_3})^{2/3}$. The activation energy $(kJ/mol)$ for the overall reaction is:
Difficult
View SolutionIf the rate of a reaction increases by $27$ times when the temperature is increased by $30 \, K$,what is the temperature coefficient of the reaction?
Medium
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