(A) The given Arrhenius equation is $\ln k = 5.0 - \frac{12000}{T}$.Comparing this with the standard Arrhenius equation $\ln k = \ln A - \frac{E_a}{RT

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(A) The given Arrhenius equation is $\ln k = 5.0 - \frac{12000}{T}$.Comparing this with the standard Arrhenius equation $\ln k = \ln A - \frac{E_a}{RT}$,we get $\frac{E_a}{R} = 12000 \ K$.Given the gas constant $R = 2 \ cal \ K^{-1} \ mol^{-1} = 2 \times 10^{-3} \ kcal \ K^{-1} \ mol^{-1}$.Therefore,$E_a = 12000 \times R = 12000 \times 2 \times 10^{-3} \ kcal \ mol^{-1}$.$E_a = 24 \ kcal \ mol^{-1}$.

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

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The specific rate constant of decomposition of a compound is given by $\ln k = 5.0 - \frac{12000}{T}$. The activation energy of decomposition for this compound at $300 \ K$ is

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A
$24 \ kcal \ mol^{-1}$
B
$12 \ kcal \ mol^{-1}$
C
$24 \ cal \ mol^{-1}$
D
$12 \ cal \ mol^{-1}$

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Class 12

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

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Consider the following graph of the kinetic energy distribution among molecules at temperature $T$. If the temperature is increased,how would the resulting graph differ from the one above?

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The rate of a chemical reaction doubles for every $10\,^{\circ}C$ rise of temperature. If the temperature is raised by $50\,^{\circ}C,$ the rate of the reaction increases by about ......... times.

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The specific rate constant of decomposition of a compound is given by $\ln k = 5.0 - \frac{12000}{T}$. The activation energy of decomposition for this compound at $300 \ K$ is

Similar Questions

The activation energy of a chemical reaction can be determined by:

When a biochemical reaction is carried out in a laboratory without the presence of enzymes,the rate of reaction is $10^{-6}$ times slower. What will be the activation energy $(E_a)$ in the presence of enzymes?

$A$ catalyst lowers the activation energy for a certain reaction from $83.314 \, kJ \, mol^{-1}$ to $75 \, kJ \, mol^{-1}$ at $500 \, K$. What will be the rate of reaction as compared to the uncatalysed reaction? Assume other things being equal.

Consider the following graph of the kinetic energy distribution among molecules at temperature $T$. If the temperature is increased,how would the resulting graph differ from the one above?

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