The activation energy for the forward reaction is $150 \ kJ \ mol^{-1}$ and that of the reverse reaction is $260 \ kJ \ mol^{-1}$. The enthalpy change for the reaction is:

How does the rate of reaction change with an increase in temperature?

The rate constant for the first order decomposition of a certain reaction is described by the equation $\ln k (s^{-1}) = 14.34 - \frac{1.25 \times 10^{4} \ K}{T}$. The energy of activation for this reaction is

The rate of a reaction:

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 $\qquad$ $[K = \text{equilibrium constant of the reaction}, k_f = \text{rate constant of forward reaction}, k_b = \text{rate constant of backward reaction}]$

The rate constant of a reaction is given by $k = P Z e^{-E_{a} / R T}$ under standard notation. In order to speed up the reaction,which of the following factors has to be decreased?