The first order rate constant for a certain reaction increases from $1.667 \times 10^{-6} \ s^{-1}$ at $727 \ ^oC$ to $1.667 \times 10^{-4} \ s^{-1}$ at $1571 \ ^oC$. The rate constant at $1150 \ ^oC$,assuming constancy of activation energy over the given temperature range is [Given : $log \ 19.9 = 1.299$ ]

The time required for $10 \%$ completion of a first order reaction at $298 \,K$ is equal to that required for its $25 \%$ completion at $308 \,K$. If the value of $A$ is $4 \times 10^{10} \,s^{-1}$,calculate $k$ at $318 \,K$ and $E_a$.

For the reaction,$aA + bB \rightarrow cC + dD$,the plot of $\log k$ vs $\frac{1}{T}$ is given below. The temperature at which the rate constant of the reaction is $10^{-4} \ s^{-1}$ is ............... $K$. (Rounded-off to the nearest integer) [Given: The rate constant of the reaction is $10^{-5} \ s^{-1}$ at $500 \ K$.]

Catalyst $A$ reduces the activation energy for a reaction by $10 \ kJ \ mol^{-1}$ at $300 \ K$. The ratio of rate $\frac{k_{T, \text{Catalysed}}}{k_{T, \text{Uncatalysed}}}$ is $e^{x}$. Find the value of $x$ [nearest integer].
[Assume that the pre-exponential factor is same in both the cases.
Given $R = 8.31 \ J \ K^{-1} \ mol^{-1}$]

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

Which one of the following is true for an exothermic reaction $A \rightleftharpoons B$,if $E_f$ and $E_b$ are the activation energies of forward and backward reactions respectively?