Question

A plot of $\text{log}{t}_{\frac{1}{2}}$ versus $\text{log}{C}_0$ is given below:


The conclusion that can be drawn from this graph is:

• A. Order = 1, $t_{\frac{1}{2}}=\frac{1}{k}$
• B. Order = 1, $t_{\frac{1}{2}}=\frac{2.303}{k}\text{log 2}$
• C. Order = 0, $t_{\frac{1}{2}}=\frac{1}{2k}$
• D. Order = 2, $t_{\frac{1}{2}}=\frac{1}{a}$

Answer 1

The correct option is B Order = 1, $t_{\frac{1}{2}}=\frac{2.303}{k}\text{log 2}$

$t_{\frac{1}{2}}\propto (C_0{)}^{1-n}$ or $t_{\frac{1}{2}}=k{C}_0^{1-n}$
(k = rate constant)
$\therefore \text{log}{t}_{\frac{1}{2}}=\text{log k}+(1-n)\text{log}{C}_0$
$y=mx+c$
Thus, plot of $\text{log}{t}_{\frac{1}{2}}$ Vs $\text{log}{C}_0$ will be linear with slope $=1-n$. But as the graph is parallel to $\text{log}{C}_0$ axis, slope = 0.
i.e., $1-n=0$ or $n=1$. Hence, it is a first order reaction and for $1^{st}$ order reactions, $t_{\frac{1}{2}}=\frac{2.303}{k}\text{log 2}$