(B) The Arrhenius equation is given by $k = A \cdot e^{-E_a / (RT)}$.As $T \rightarrow \infty$,the exponent term $-E_a / (RT) \rightarrow 0$.Therefore

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(B) The Arrhenius equation is given by $k = A \cdot e^{-E_a / (RT)}$.As $T \rightarrow \infty$,the exponent term $-E_a / (RT) \rightarrow 0$.Therefore,$e^{-E_a / (RT)} \rightarrow e^0 = 1$.Thus,the rate constant $k$ approaches the Arrhenius factor $A$.Given $A = 6.0 \times 10^{14} \, s^{-1}$,the value of the rate constant as $T \rightarrow \infty$ is $6.0 \times 10^{14} \, s^{-1}$.

Class 12 Chemistry Chemical Kinetics Collision theory, Energy of activation and Arrhenius equation

Medium

For a chemical reaction,the rate constant,activation energy,and Arrhenius factor at $25 \, ^\circ C$ are $3.0 \times 10^{-4} \, s^{-1}$,$104.4 \, kJ \, mol^{-1}$,and $6.0 \times 10^{14} \, s^{-1}$ respectively. Find the value of the rate constant as $T \rightarrow \infty$.

A
$2.0 \times 10^{18} \, s^{-1}$
B
$6.0 \times 10^{14} \, s^{-1}$
C
Infinity
D
$3.6 \times 10^{30} \, s^{-1}$

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Class 12

Chemistry Questions

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

For a chemical reaction at $27^{\circ} C$, the activation energy is $600 R$. The ratio of the rate constants at $327^{\circ} C$ to that of at $27^{\circ} C$ will be

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For a first-order gaseous reaction,a plot of $\log \, k$ versus $1/T$ gives a straight line with a slope of $-8000$. Calculate the activation energy $(E_a)$ of the reaction in $cal$.

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The increase in rate constant of a chemical reaction with increasing temperature is due to the fact$(s)$ that:

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At $300 \ K$,the activation energy and rate constant for the given reaction are $10 \ kJ \ mol^{-1}$ and $2.4 \times 10^{-5} \ s^{-1}$ respectively. At what temperature (in $K$) will the value of $t_{1/2}$ be $2 \ hr$?
$2N_2O_5 \to 4NO_2 + O_2$

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The minimum energy required for molecules to participate in a chemical reaction is called $.......$.

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For a chemical reaction,the rate constant,activation energy,and Arrhenius factor at $25 \, ^\circ C$ are $3.0 \times 10^{-4} \, s^{-1}$,$104.4 \, kJ \, mol^{-1}$,and $6.0 \times 10^{14} \, s^{-1}$ respectively. Find the value of the rate constant as $T \rightarrow \infty$.

Similar Questions

For a chemical reaction at $27^{\circ} C$, the activation energy is $600 R$. The ratio of the rate constants at $327^{\circ} C$ to that of at $27^{\circ} C$ will be

For a first-order gaseous reaction,a plot of $\log \, k$ versus $1/T$ gives a straight line with a slope of $-8000$. Calculate the activation energy $(E_a)$ of the reaction in $cal$.

The increase in rate constant of a chemical reaction with increasing temperature is due to the fact$(s)$ that:

At $300 \ K$,the activation energy and rate constant for the given reaction are $10 \ kJ \ mol^{-1}$ and $2.4 \times 10^{-5} \ s^{-1}$ respectively. At what temperature (in $K$) will the value of $t_{1/2}$ be $2 \ hr$?$2N_2O_5 \to 4NO_2 + O_2$

The minimum energy required for molecules to participate in a chemical reaction is called $.......$.

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At $300 \ K$,the activation energy and rate constant for the given reaction are $10 \ kJ \ mol^{-1}$ and $2.4 \times 10^{-5} \ s^{-1}$ respectively. At what temperature (in $K$) will the value of $t_{1/2}$ be $2 \ hr$?
$2N_2O_5 \to 4NO_2 + O_2$