For this reaction:
\[K = \frac{2.303}{t} \log \frac{[A]_0}{[A]}\]
Given: \( k = 4.6 \times 10^{-2} \, \text{s}^{-1}, \, [A]_0 = 1 \, \text{M}, \, [A] = 0.1 \, \text{M} \)
\[4.6 \times 10^{-2} = \frac{2.303}{t} \log \frac{1}{0.1}\]
\[4.6 \times 10^{-2} = \frac{2.303}{t} \times 1\]
\[t = \frac{2.303}{4.6 \times 10^{-2}} \approx 50 \, \text{sec.}\]
A(g) $ \rightarrow $ B(g) + C(g) is a first order reaction.
The reaction was started with reactant A only. Which of the following expression is correct for rate constant k ?
Rate law for a reaction between $A$ and $B$ is given by $\mathrm{R}=\mathrm{k}[\mathrm{A}]^{\mathrm{n}}[\mathrm{B}]^{\mathrm{m}}$. If concentration of A is doubled and concentration of B is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction $\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)$ is
For $\mathrm{A}_{2}+\mathrm{B}_{2} \rightleftharpoons 2 \mathrm{AB}$ $\mathrm{E}_{\mathrm{a}}$ for forward and backward reaction are 180 and $200 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. If catalyst lowers $\mathrm{E}_{\mathrm{a}}$ for both reaction by $100 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Which of the following statement is correct?
Match List-I with List-II: List-I