On increasing the temperature,the rate of the reaction increases because of

  • A
    Decrease in the number of collisions
  • B
    Decrease in the energy of activation
  • C
    Decrease in the number of activated molecules
  • D
    Increase in the number of effective collisions

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The rate constant for the decomposition of $N_{2}O_{5}$ at various temperatures is given below:
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Draw a graph between $\ln k$ and $1 / T$ and calculate the values of $A$ and $E_{a}.$ Predict the rate constant at $30^{\circ}C$ and $50^{\circ}C$.

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For reaction $A \to B$,the rate constant $k_1 = A_1 e^{-E_{a_1} / (RT)}$ and for the reaction $X \to Y$ the rate constant $k_2 = A_2 e^{-E_{a_2} / (RT)}$. If $A_1 = 10^8$,$A_2 = 10^{10}$ and $E_{a_1} = 600 \ cal/mol$,$E_{a_2} = 1800 \ cal/mol$,then the temperature at which $k_1 = k_2$ is (Given : $R = 2 \ cal/K \cdot mol$)

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