Question

$Zn +2 H ^{\oplus} \rightarrow Zn ^{2+}+ H _2$
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. $\frac{d x}{d t}=k[ Zn ]^0\left[ H ^{\oplus}\right]^2$
• B. $\frac{d x}{d t}=k[ Zn ]\left[ H ^{\oplus}\right]^2$
• 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 $pH =3]$
$r_2=100 r_1 \propto\left[10^{-2}\right]^x [$ 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 $=\left(\frac{d x}{d t}\right)=k[ Zn ]\left[ H ^{\oplus}\right]^2$
So (b) is the correct answer
c. $\left(\frac{d x}{d t}\right)=r_1=k[ Zn ]\left[ H ^{\oplus}\right]^2$
$r_2=k[ Zn ]\left[\frac{ H ^{\oplus}}{2}\right]^2$
Thus, $r_2=r_1$
d. $ r_1=k[ Zn ]\left[ H ^{\oplus}\right]^2 \ldots \text { (i) } $
$r_2=k[ Zn ]\left[2 H ^{\oplus}\right]^2 \ldots( ii )$
Divide (ii) by (i),
$r_2 = 4r_1$