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

Answer 1

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

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

i. Since $r_{1/2}$ is independent of concentration of $Zn$ at constant $pH$ means that the order w.r.t $\left[Zn\right]=1$.

ii. Let $r_1\propto [H^{\oplus }{]}^x\Rightarrow r_1\propto [{10}^{-3}{]}^x[whenpH=3]$

$r_2=100r_1\propto \left[{10}^{-2}{]}^x\right[whenpH=2]$

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 = $\left(\frac{dx}{dt}\right)=k\left[Zn\right][H^{\oplus }{]}^2$

So (b) is the correct answer.

c. $\left(\frac{dx}{dt}\right)=r_1=k\left[Zn\right][H^{\oplus }{]}^2$

$r_2=k\left[Zn\right]{\left[\frac{H^{\oplus }}{2}\right]}^2$

Thus, $r_2=r_1$

d. $r_1=k\left[Zn\right][H^{\oplus }{]}^2$

$r_2=k\left[Zn\right][2H^{\oplus }{]}^2$

Divide (ii) by (i)

$r_2=4r_1$