(A) The Arrhenius equation is given by $\ln k = \ln A - \frac{E_{a}}{RT}$.Comparing this with the equation of a straight line $y = mx + c$,where $y =

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(A) The Arrhenius equation is given by $\ln k = \ln A - \frac{E_{a}}{RT}$.Comparing this with the equation of a straight line $y = mx + c$,where $y = \ln k$ and $x = \frac{1}{T}$,the slope $m = -\frac{E_{a}}{R}$.Given that the slope is $-5 \times 10^{3} \, K$,we have $-\frac{E_{a}}{R} = -5 \times 10^{3} \, K$.Therefore,$E_{a} = 5 \times 10^{3} \times 8.314 \, J \, mol^{-1} = 41570 \, J \, mol^{-1}$.Converting to $kJ \, mol^{-1}$,we get $E_{a} = 41.57 \, kJ \, mol^{-1} \approx 41.5 \, kJ \, mol^{-1}$.

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

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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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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

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The rate constant is doubled when temperature increases from $27^{\circ} C$ to $37^{\circ} C$. What is the activation energy in $kJ \ mol^{-1}$?

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What is the activation energy $(kJ \, mol^{-1})$ for a reaction if its rate constant doubles when the temperature is raised from $300 \, K$ to $400 \, K$ ? $(R = 8.314 \, J \, mol^{-1} \, K^{-1})$

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For an exothermic reaction $X \rightarrow Y$,the activation energy is $30 \ kJ \ mol^{-1}$. If the enthalpy change $(\Delta H)$ for the reaction is $-20 \ kJ \ mol^{-1}$,then the activation energy for the reverse reaction is . . . . . . $kJ \ mol^{-1}$.

Medium

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Fill in the blanks:
$1.$ Common collision means ......... phenomenon in species.
$2.$ In collision theory,activation energy and proper orientation of the molecules together determine the ......... for an effective collision.

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Activation energy is defined as:

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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 $kJ \, mol^{-1}$)$[\text{Given } R = 8.314 \, J \, K^{-1} \, mol^{-1}]$

Similar Questions

The rate constant is doubled when temperature increases from $27^{\circ} C$ to $37^{\circ} C$. What is the activation energy in $kJ \ mol^{-1}$?

What is the activation energy $(kJ \, mol^{-1})$ for a reaction if its rate constant doubles when the temperature is raised from $300 \, K$ to $400 \, K$ ? $(R = 8.314 \, J \, mol^{-1} \, K^{-1})$

For an exothermic reaction $X \rightarrow Y$,the activation energy is $30 \ kJ \ mol^{-1}$. If the enthalpy change $(\Delta H)$ for the reaction is $-20 \ kJ \ mol^{-1}$,then the activation energy for the reverse reaction is . . . . . . $kJ \ mol^{-1}$.

Fill in the blanks:$1.$ Common collision means ......... phenomenon in species.$2.$ In collision theory,activation energy and proper orientation of the molecules together determine the ......... for an effective collision.

Activation energy is defined as:

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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 $kJ \, mol^{-1}$)
$[\text{Given } R = 8.314 \, J \, K^{-1} \, mol^{-1}]$

Fill in the blanks:
$1.$ Common collision means ......... phenomenon in species.
$2.$ In collision theory,activation energy and proper orientation of the molecules together determine the ......... for an effective collision.