The reaction between A2 (g) and B2 (g) was carried out in a sealed isothermal container. The rate law for the reaction was found to be:
Rate = \( k[\text{A}_2][\text{B}_2] \)
If 1 mole of A2 (g) was added to the reaction chamber and the temperature was kept constant, then predict the change in rate of the reaction and the rate constant.
According to the rate law:
Rate = \( k[\text{A}_2][\text{B}_2] \)
When 1 mole of A2 is added, [A2] increases. Since the temperature remains constant, the rate constant \( k \) remains unchanged.
Let the initial concentration of A2 be [A2], then after adding 1 mole, the new concentration increases, thus increasing the rate proportionally.
Conclusion: The rate of reaction will increase due to increased [A2], but the rate constant \( k \) remains the same.
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?