Question

$Zn+2H^{+}\to Z{n}^{2+}+H_2$

Half – life period is independent of concentration of constant pH. For the constant concentration of Zn, rate becomes 100 times when pH is decreased from 3 to 2. Hence,

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

Answer 1

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

The correct options are
B

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

C

rate is not affected if concentration of zinc is made four times and that of $H^{+}$ ion is halved

D

rate becomes four times if concentration of $H^{+}$ ion is doubled at constant Zn concentration.

Half-life period independent of conc. of Zn then order w.r.t Zn is 1st

Rate = $k\left[Zn\right][H^{+}{]}^n$we can write

We are increasing the $\left[H^{+}\right]$ ion concentration by changing the pH from 3 to 2. So the concentration of $\left[H^{+}\right]$ ions becomes ${10}^{-2}$ from ${10}^{-3}$, i.e. increases by a factor of 10. When we increase the concentration of $\left[H^{+}\right]$ by 10 times, reaction rate increases by 100 times = ${10}^2$. Hence, order of reaction wrt $\left[H^{+}\right]$ is 2.

n = 2

Then Rate = $k\left[Zn\right][H^{+}{]}^2$

(c) Conc. of Zn four times and $H^{+}$ ion is halved.

${\text{Rate}}_2=k\times 4\times \left[Zn\right]{\frac{\left[H^{+}\right]}{2^2}}^2$

${\text{Rate}}_2=k\left[Zn\right][H^{+}{]}^2={\text{Rate}}_1$

(d) Similarly if conc. of $H^{+}$ doubled and [Zn] constant

${\text{Rate}}_2=k\left[Zn\right]\times 4\times [H^{+}{]}^2$

${\text{Rate}}_2=4\times {\text{Rate}}_1$