JEE MAIN - Chemistry (2003 - No. 41)
For the reaction system:
2NO(g) + O2(g) $$\to$$ 2NO2(g) volume is suddenly reduce to half its value by increasing the pressure on it. If the reaction is of first order with respect to O2 and second order with respect to NO, the rate of reaction will
2NO(g) + O2(g) $$\to$$ 2NO2(g) volume is suddenly reduce to half its value by increasing the pressure on it. If the reaction is of first order with respect to O2 and second order with respect to NO, the rate of reaction will
diminish to one-eighth of its initial value
increase to eight times of its initial value
increase to four times of its initial value
diminish to one-fourth of its initial value
Explanation
$$r = k\left[ {{O_2}} \right]{\left[ {NO} \right]^2}.$$
When the volume is reduced to $$1/2,$$ the conc. will double
$$\therefore$$ $$\,\,\,\,\,$$ New rate $$ = k\left[ {2{O_2}} \right]{\left[ {2NO} \right]^2} = 8k\left[ {{O^2}} \right]{\left[ {NO} \right]^2}$$
The new rate increases to eight times of its initial.
When the volume is reduced to $$1/2,$$ the conc. will double
$$\therefore$$ $$\,\,\,\,\,$$ New rate $$ = k\left[ {2{O_2}} \right]{\left[ {2NO} \right]^2} = 8k\left[ {{O^2}} \right]{\left[ {NO} \right]^2}$$
The new rate increases to eight times of its initial.