On increasing the temperature,the rate of the reaction increases because of
- ADecrease in the number of collisions
- BDecrease in the energy of activation
- CDecrease in the number of activated molecules
- DIncrease in the number of effective collisions
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Similar Questions
For a reaction,given below is the graph of $\ln k$ vs $\frac{1}{T}$. The activation energy for the reaction is equal to $...... \ cal \ mol^{-1}$. (Nearest integer). (Given : $R = 2 \ cal \ K^{-1} \ mol^{-1}$)
In a reaction,the rate of reaction doubles with a $10^\circ C$ rise in temperature. If the temperature is increased from $10^\circ C$ to $100^\circ C$,by how many times will the rate of reaction increase?
Easy
View Solution$A$ reaction takes place in three steps with individual rate constant and activation energy,
Step Rate constant and Activation energy $Step \ 1$ $k_1, E_{a_1} = 180 \ kJ \ mol^{-1}$ $Step \ 2$ $k_2, E_{a_2} = 80 \ kJ \ mol^{-1}$ $Step \ 3$ $k_3, E_{a_3} = 50 \ kJ \ mol^{-1}$
Overall rate constant,$k = (k_1 k_2 / k_3)^{2/3}$. The overall activation energy of the reaction will be ........ $kJ \ mol^{-1}$.
| Step | Rate constant and Activation energy |
| $Step \ 1$ | $k_1, E_{a_1} = 180 \ kJ \ mol^{-1}$ |
| $Step \ 2$ | $k_2, E_{a_2} = 80 \ kJ \ mol^{-1}$ |
| $Step \ 3$ | $k_3, E_{a_3} = 50 \ kJ \ mol^{-1}$ |
Overall rate constant,$k = (k_1 k_2 / k_3)^{2/3}$. The overall activation energy of the reaction will be ........ $kJ \ mol^{-1}$.
Difficult
View SolutionRegarding the velocity in a reversible reaction,which is the correct explanation of the effect of a catalyst?
Easy
View SolutionFor a reversible reaction $A \rightleftharpoons B$,which one of the following statements is wrong from the given energy profile diagram?
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