Calculate the half-life of a first order reaction from their rate constants given below:
\((i)\) \(Half \ life\) \(t_{\frac 12} = \frac {0.693}{k}\)
= \(\frac {0.693}{200\ s^{-1}} \)
= \(3.47 \times 10^{-3} s \ (approximately)\)
\((ii)\) \(Half \ life\) \(t_{\frac 12} = \frac {0.693}{k}\)
=\(\frac { 0.693}{2 \ min^{-1}}\)
= \(0.35 \ min \ (approximately)\)
\((iii)\) \(Half \ life\) \(t_{\frac 12} = \frac {0.693}{k}\)
= \(\frac {0.693}{4 \ year^{-1}}\)
= \(0.173\ years\ (approximately)\)
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.