Chemistry - Chemical Kinetics Question with Solution | TestHub

Key Idea : By comparing the rate and concentration the order of the reaction can be calculated Let the rate of the reaction wrt $\left[{\mathrm{CH}}_3{\mathrm{COCH}}_3\right]$, $\left[{\mathrm{Br}}_2\right]$ and $\left[{\mathrm{H}}^{+}\right]$are $x,y$ and $z$ respectively. Thus, $\text{Rate }\propto {\left[{\mathrm{CH}}_3{\mathrm{COCH}}_3\right]}^x{\left[{\mathrm{Br}}_2\right]}^y{\left[{\mathrm{H}}^{+}\right]}^z$ $5.7\times 10^{-5}=[0.30{]}^x[0.05{]}^y[0.05{]}^z$ $5.7\times 10^{-5}=[0.30{]}^x(0.10{)}^y(0.05{)}^z$ $1.2\times 10^{-4}=[0.30{]}^x(0.10{)}^y(0.10{)}^z$ $3.1\times 10^{-4}=[0.40{]}^x(0.05{)}^y(0.20{)}^z$ From Eqs (i) and (ii) $y=0$ From Eqs (ii) and (iii) $z=1$ From Eqs (i) and (iv) $x=1$ Thus, rate law $\propto \left[{\mathrm{CH}}_3{\mathrm{COCH}}_3\right]\left[{\mathrm{H}}^{+}\right]$ $=k\left[{\mathrm{CH}}_3{\mathrm{COCH}}_3\right]\left[{\mathrm{H}}^{+}\right]$