$A$ catalyst

  • A
    changes the equilibrium constant
  • B
    lowers the activation energy
  • C
    increases the forward and backward reactions at different speeds.
  • D
    follows same mechanism for the reaction.

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Similar Questions

For a first order reaction,a plot of $\ln k$ ($y$-axis) and $\frac{1}{T}$ ($x$-axis) gave a straight line with a slope equal to $-10^3 \ K$ and an intercept equal to $2.303$ (on $y$-axis). What is the activation energy ($E_a$ in $kJ \ mol^{-1}$) of the reaction? (Given $R = 8.314 \ J \ mol^{-1} \ K^{-1}$)

Explain the rate of reaction with energy of activation and temperature with the help of the Arrhenius equation and state its importance.

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The plot of $\log k_f$ versus $1 / T$ for a reversible reaction $A_{(g)} \rightleftharpoons P_{(g)}$ is shown. Pre-exponential factors for the forward and backward reactions are $10^{15} \ s^{-1}$ and $10^{11} \ s^{-1}$,respectively. If the value of $\log K$ for the reaction at $500 \ K$ is $6$,the value of $|\log k_b|$ at $250 \ K$ is $\qquad$ $[K = \text{equilibrium constant of the reaction}, k_f = \text{rate constant of forward reaction}, k_b = \text{rate constant of backward reaction}]$

According to Arrhenius theory,the activation energy is

$A$ reaction takes place in three steps with individual rate constant and activation energy as follows:
$Step$$Rate \ constant$$Activation \ energy$
$Step-1$$k_1$$E_{a1} = 180 \ kJ/mol$
$Step-2$$k_2$$E_{a2} = 80 \ kJ/mol$
$Step-3$$k_3$$E_{a3} = 50 \ kJ/mol$
If overall rate constant,$k = (\frac{k_1 k_2}{k_3})^{2/3}$,then overall activation energy of the reaction will be .......... $kJ/mol$.

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