The rate law of the reaction $2N_2O_5 \to 4NO_2 + O_2$ is

$100 \text{ cm}^3$ of $1 \text{ M } CH_3COOH$ was mixed with $100 \text{ cm}^3$ of $2 \text{ M } CH_3OH$ to form an ester. The change in the initial rate if each solution is diluted with an equal volume of water would be: (in $\text{times}$)

For the reaction $Cl_{2(aq)} + H_2S_{(aq)} \to S_{(s)} + 2H^+_{(aq)} + 2Cl^-_{(aq)}$, the rate law is given by $\text{Rate} = K[Cl_2][H_2S]$. Which of the following mechanisms is consistent with this rate law?
$(A)$ $Cl_2 + H_2S \to H^+ + Cl^- + Cl^+ + HS^-$ (slow); $Cl^+ + HS^- \to H^+ + Cl^- + S$ (fast)
$(B)$ $H_2S \rightleftharpoons H^+ + HS^-$ (fast equilibrium); $Cl_2 + HS^- \to 2Cl^- + H^+ + S$ (slow)

In the reaction $A + B \to \text{Products}$,if $B$ is taken in excess,then it is an example of

When the temperature increases from $300 \, K$ to $310 \, K$,the rate of reaction increases by $2.3$ times. If the rate constant at $300 \, K$ is $x$,then the rate constant at $310 \, K$ will be equal to .......

The rate constant of esterification is given by $k = k^{\prime} [H_2O]$. If the rate constant $k = 2.0 \times 10^{-3} \ min^{-1}$,calculate $k^{\prime}$. (Assume $[H_2O] = 55.5 \ mol \ L^{-1}$)