For a reaction,the graph of $\ln k$ (on y-axis) and $1 / T$ (on x-axis) is a straight line with a slope $-2 \times 10^4 \ K$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is $(R = 8.3 \ J \ K^{-1} \ mol^{-1})$

The first order rate constant for the decomposition of $CaCO_3$ at $700 \ K$ is $6.36 \times 10^{-3} \ s^{-1}$ and activation energy is $209 \ kJ \ mol^{-1}$. Its rate constant (in $s^{-1}$) at $500 \ K$ is $x \times 10^{-6}$. The value of $x$ is ..... (Nearest integer)
Given $R=8.31 \ J \ K^{-1} \ mol^{-1} ; \log(6.36 \times 10^{-3})=-2.19 ; [10^{-4.79}=1.62 \times 10^{-5}]$

Fill in the blanks:
$(a)$ $A$ catalyst ......... the change of equilibrium in reaction.
$(b)$ Boltzmann and Maxwell used ......... to explain the rate of reaction.

The rate constant for the decomposition of a hydrocarbon is given as $K = (4.5 \times 10^{11} \ s^{-1}) e^{-(28000 \ K) / T}$. The activation energy of the reaction (in $J/mol$) is:

At $T \ K$,the following equation is obtained for a first order reaction: $\log \frac{k}{A} = -\frac{x}{T}$. The activation energy for this reaction is equal to $(R = \text{gas constant})$

When the temperature of a reaction is increased from $300 \ K$ to $310 \ K$,the rate of the reaction increases by $2.5$ times. If the rate constant of the reaction at $300 \ K$ is $K$,what will be the rate constant at $310 \ K$?