
The graph shows a linear relationship between log |R₀| and time. This suggests a first-order reaction.
In a first-order reaction, the concentration of the reactant decreases exponentially over time. This means:
\[ \ln [R] = \ln [R_0] - kt \]
Taking the logarithm (base 10) of both sides also gives a linear equation, which justifies that a plot of \(\log [R]\) versus time is a straight line.
For a first-order reaction, the slope of the \(\log [R]\) vs time graph is equal to \(-k\), where \(k\) is the rate constant.
So, the slope \(m\) of the graph is:
\[ \text{slope} = -k \]
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?