Question

$Zn+2H^{\oplus }\to Z{n}^{2+}+H_2$
The half-life period is independent of the concentration of zinc at constant $pH$. For the constant concentration of $Zn$, the rate becomes $100$ times when $pH$ is decreased from $3$ to $2$ . Hence,

• A. $\frac{dx}{dt}=k[Zn{]}^0{\left[H^{\oplus }\right]}^2$
• B. $\frac{dx}{dt}=k[Zn]{\left[H^{\oplus }\right]}^2$
• C. Rate is not affected if the concentration of zinc is made four times and that of $H^{\oplus }$ ion is halved
• D. Rate becomes four times if the concentration of $H^{\oplus }$ ion is doubled at constant $Zn$ concentration

Answer 1

Answer: (B), (C), (D)

1. Since $t_{1/2}$ is independent of concentration of $Zn$ at constant $pH$ means that the order w.r.t. $[Zn]=1$
2. Let $r_1\propto {\left[H^{\oplus }\right]}^x\Rightarrow r_1\propto {\left[10^{-3}\right]}^x$ [when $pH=3]$
$r_2=100r_1\propto {\left[10^{-2}\right]}^x[$ when $pH=2]$
Thus, $\frac{r_2}{r_1}={\left[\frac{10^{-2}}{10^{-3}}\right]}^x$
$100=(10{)}^2=[10{]}^x\Rightarrow x=2$
Hence, order w.r.t. $\left[H^{\oplus }\right]=2$
b. Hence, the rate $=\left(\frac{dx}{dt}\right)=k[Zn]{\left[H^{\oplus }\right]}^2$
So (b) is the correct answer
c. $\left(\frac{dx}{dt}\right)=r_1=k[Zn]{\left[H^{\oplus }\right]}^2$
$r_2=k[Zn]{\left[\frac{H^{\oplus }}{2}\right]}^2$
Thus, $r_2=r_1$
d. $r_1=k[Zn]{\left[H^{\oplus }\right]}^2\dots \text{ (i) }$
$r_2=k[Zn]{\left[2H^{\oplus }\right]}^2\dots (ii)$
Divide (ii) by (i),
$r_2=4r_1$