The rate of formation of B is:
\[ \frac{d[\text{B}]}{dt} = k_1[\text{A}] - k_2[\text{B}]. \]
For the rate of formation of B to be zero:
\[ \frac{d[\text{B}]}{dt} = 0. \]
Substitute:
\[ k_1[\text{A}] - k_2[\text{B}] = 0. \]
Rearrange to find [B]:
\[ k_1[\text{A}] = k_2[\text{B}] \implies [\text{B}] = \frac{k_1}{k_2}[\text{A}]. \]
Thus, the concentration of B is:
\[ [\text{B}] = \left(\frac{k_1}{k_2}\right)[\text{A}]. \]
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