Question

The rate of a reaction starting with initial concentration of $2\times {10}^{-3}M$ and $1\times {10}^{-3}M$ are equal to $240\times {10}^{-4}M{s}^{-1}$ and $0.60\times {10}^{4}Mse{c}^{-1}$ respectively. The rate constant is:

• A. $60mo{l}^{-1}Lse{c}^{-1}$
• B. $30mo{l}^{-1}Lse{c}^1$
• C. $20mo{l}^{1}Lse{c}^{-1}$
• D. None of these

Answer 1

The correct option is A $60mo{l}^{-1}Lse{c}^{-1}$

$(r_0{)}_1=K[A_0{]}_1^a$

$(r_0{)}_2=K[A_0{]}_2^a$

$\therefore \frac{(r_0{)}_1}{(r_0{)}_2}={\left\{\frac{[A_0{]}_1}{[A_0{]}_2}\right\}}^a$

or $a=\frac{log(r_0{)}_1-log(r_0{)}_2}{log[A_0{]}_1-log[A_0{]}_2}$

$=\frac{log[2.40\times {10}^{-4}]-log[0.60\times {10}^{-4}]}{log[2\times {10}^{-3}]-log[1\times {10}^{-3}]}$

$=\frac{-3.62+4.22}{-2.70+3}=2$

$\therefore r=K[A{]}^2$ i.e., II order

Also, $K=\frac{r}{[A{]}^2}=\frac{240\times {10}^{-4}}{[2\times {10}^{-3}{]}^2}=60mo{l}^{-1}Lse{c}^{-1}$

Hence, option A is correct.