Let the rate law be given by: Rate = k[A]x[B]y, where x and y are the orders of the reaction with respect to A and B respectively.
From the given data:
Comparing the first and second rows (keeping [B] constant):
\[\frac{4\times10^{-3}}{2\times10^{-3}}=\frac{k[0.2]^{x}[0.1]^{y}}{k[0.1]^{x}[0.1]^{y}}\]
\[2=2^{x}\]
\[x=1\]
Comparing the second and third rows (keeping [A] constant):
\[\frac{1.6\times10^{-2}}{4\times10^{-3}}=\frac{k[0.2]^{x}[0.2]^{y}}{k[0.2]^{x}[0.1]^{y}}\]
\[4=2^{y}\]
\[y=2\]
Therefore, the order of the reaction with respect to A is 1, and with respect to B is 2.
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?
List I | List II | ||
---|---|---|---|
A | Mesozoic Era | I | Lower invertebrates |
B | Proterozoic Era | II | Fish & Amphibia |
C | Cenozoic Era | III | Birds & Reptiles |
D | Paleozoic Era | IV | Mammals |