A first order reaction takes 40 min for 30% decomposition. Calculate \(t_{\frac 12}\).
For a first order reaction,
\(t = \frac {2.303}{k} log\ \frac { [R]_0}{[R]}\)
\(k= \frac {2.303}{40 \ min} log \frac {100}{100-30}\)
\(k= \frac {2.303}{40 \ min} log \frac {10}{7}\)
\(k = 8.918 \times10^{-3} min^{-1}\)
Therefore, \(t_{\frac 12}\) of the decomposition reaction is
\(t_{\frac 12}= \frac {0.693}{k}\)
\(t_{\frac 12} = \frac {0.693}{8.918 \times 10^{-3}} min\)
\(t_{\frac 12 }= 77.7 \ min \ (approximately)\)
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?
The Order of reaction refers to the relationship between the rate of a chemical reaction and the concentration of the species taking part in it. In order to obtain the reaction order, the rate equation of the reaction will given in the question.