(D) The Arrhenius equation is given by $\log k = \log A - \frac{E_a}{2.303 \ RT}$.Comparing this with the given equation $\log k = 7.14 - \frac{1 \tim

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(D) The Arrhenius equation is given by $\log k = \log A - \frac{E_a}{2.303 \ RT}$.Comparing this with the given equation $\log k = 7.14 - \frac{1 \times 10^4 \ K}{T}$,we get $\frac{E_a}{2.303 \ R} = 1 \times 10^4 \ K$.Therefore,$E_a = 2.303 \times R \times 10^4 \ K$.Substituting $R = 8.3 \ J \ K^{-1} \ mol^{-1}$,we get $E_a = 2.303 \times 8.3 \times 10^4 \ J \ mol^{-1} = 191.149 \times 10^3 \ J \ mol^{-1} = 191.149 \ kJ \ mol^{-1}$.Rounding to one decimal place,the activation energy is $191.1 \ kJ \ mol^{-1}$.

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

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For a first order decomposition of a certain reaction,rate constant is given by the equation $\log k \left( s^{-1} \right) = 7.14 - \frac{1 \times 10^4 \ K}{T}$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is $(R = 8.3 \ J \ K^{-1} \ mol^{-1})$ (in $.1$)

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$161$
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$171$
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$181$
D
$191$

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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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If 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?

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At temperature $T \ K$,the rate constant of a reaction is $1/10$th of the rate constant at temperature $2T \ K$. What will be the activation energy of this reaction (in $RT$)?

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The rate of a certain reaction doubles when the temperature changes from $27\,^oC$ to $37\,^oC$. The activation energy for the reaction is ........... $kJ\,mol^{-1}$.

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Among the following,the $INCORRECT$ statement regarding the collision theory of chemical reaction is:

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The slope of the Arrhenius plot $(\ln k \, vs \, \frac{1}{T})$ of a first-order reaction is $-5 \times 10^{3} \, K$. The value of $E_{a}$ of the reaction is: (in $kJ \, mol^{-1}$)
$[\text{Given } R = 8.314 \, J \, K^{-1} \, mol^{-1}]$

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For a first order decomposition of a certain reaction,rate constant is given by the equation $\log k \left( s^{-1} \right) = 7.14 - \frac{1 \times 10^4 \ K}{T}$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is $(R = 8.3 \ J \ K^{-1} \ mol^{-1})$ (in $.1$)

Similar Questions

If 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?

At temperature $T \ K$,the rate constant of a reaction is $1/10$th of the rate constant at temperature $2T \ K$. What will be the activation energy of this reaction (in $RT$)?

The rate of a certain reaction doubles when the temperature changes from $27\,^oC$ to $37\,^oC$. The activation energy for the reaction is ........... $kJ\,mol^{-1}$.

Among the following,the $INCORRECT$ statement regarding the collision theory of chemical reaction is:

The slope of the Arrhenius plot $(\ln k \, vs \, \frac{1}{T})$ of a first-order reaction is $-5 \times 10^{3} \, K$. The value of $E_{a}$ of the reaction is: (in $kJ \, mol^{-1}$)$[\text{Given } R = 8.314 \, J \, K^{-1} \, mol^{-1}]$

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The slope of the Arrhenius plot $(\ln k \, vs \, \frac{1}{T})$ of a first-order reaction is $-5 \times 10^{3} \, K$. The value of $E_{a}$ of the reaction is: (in $kJ \, mol^{-1}$)
$[\text{Given } R = 8.314 \, J \, K^{-1} \, mol^{-1}]$