(D) The graph represents the Maxwell-Boltzmann distribution of kinetic energy. When the temperature increases from $T_1$ to $T_2$ $(T_2 > T_1)$,the pe

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Area $I$ would decrease and area $II$ would increase

(D) The graph represents the Maxwell-Boltzmann distribution of kinetic energy. When the temperature increases from $T_1$ to $T_2$ $(T_2 > T_1)$,the peak of the curve shifts to the right and the curve flattens. This means that the fraction of molecules with lower kinetic energy (represented by area $I$) decreases,while the fraction of molecules with higher kinetic energy (represented by area $II$) increases. Therefore,area $I$ decreases and area $II$ increases.

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

Medium

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?

A
Both area $I$ and $II$ would increase
B
Both area $I$ and $II$ would decrease
C
Area $I$ would increase and area $II$ would decrease
D
Area $I$ would decrease and area $II$ would increase

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

The following equation is obtained for a first order reaction at $300 \ K$.
$\log_{10} \frac{k}{A} = 0.00174$
What is the activation energy (in $J \ mol^{-1}$) of the reaction?
$(R = 8.314 \ J \ mol^{-1} \ K^{-1})$

MediumAP EAMCET 2025

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The velocity constant of a reaction at $290 \ K$ was found to be $3.2 \times 10^{-3}$. At $310 \ K$,it will be about:

Medium

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Assertion $(A)$ : $A$ catalyst increases the rate of a reaction.
Reason $(R)$ : In presence of a catalyst,the activation energy of the reaction increases.
The correct answer is

MediumAP EAMCET 2007

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For an endothermic reaction,where $\Delta H$ represents the enthalpy of the reaction in $kJ/mol$,the minimum value for the energy of activation will be

MediumIIT 1992

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For an endothermic reaction,$\Delta H$ represents the enthalpy of the reaction in $kJ \ mol^{-1}$. The minimum amount of activation energy will be

Easy

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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?

Similar Questions

The following equation is obtained for a first order reaction at $300 \ K$.$\log_{10} \frac{k}{A} = 0.00174$What is the activation energy (in $J \ mol^{-1}$) of the reaction?$(R = 8.314 \ J \ mol^{-1} \ K^{-1})$

The velocity constant of a reaction at $290 \ K$ was found to be $3.2 \times 10^{-3}$. At $310 \ K$,it will be about:

Assertion $(A)$ : $A$ catalyst increases the rate of a reaction.Reason $(R)$ : In presence of a catalyst,the activation energy of the reaction increases.The correct answer is

For an endothermic reaction,where $\Delta H$ represents the enthalpy of the reaction in $kJ/mol$,the minimum value for the energy of activation will be

For an endothermic reaction,$\Delta H$ represents the enthalpy of the reaction in $kJ \ mol^{-1}$. The minimum amount of activation energy will be

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The following equation is obtained for a first order reaction at $300 \ K$.
$\log_{10} \frac{k}{A} = 0.00174$
What is the activation energy (in $J \ mol^{-1}$) of the reaction?
$(R = 8.314 \ J \ mol^{-1} \ K^{-1})$

Assertion $(A)$ : $A$ catalyst increases the rate of a reaction.
Reason $(R)$ : In presence of a catalyst,the activation energy of the reaction increases.
The correct answer is