A[M] | B[M] | initial rate of formation of D | |
I | 0.1 | 0.1 | 6.0 × 10-3 |
II | 0.3 | 0.2 | 7.2 × 10-2 |
III | 0.3 | 0.4 | 2.88 × 10-1 |
IV | 0.4 | 0.1 | 2.40 × 10-2 |
From the rate law:
\[ r = k[A]^x[B]^y \]
Using the data:
\[ \text{From (I) and (IV): } \frac{2.4 \times 10^{-2}}{6 \times 10^{-3}} = \frac{(0.4)^x}{(0.1)^x} \implies 4 = 4^x \implies x = 1 \]
\[ \text{From (III) and (II): } \frac{2.88 \times 10^{-1}}{7.2 \times 10^{-2}} = \frac{(0.4)^y}{(0.2)^y} \implies 4 = 2^y \implies y = 2 \]
Overall order = \(x + y = 1 + 2 = 3\).
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