Question

For the reaction system: 2NO(g) + O2(g)2NO2(g), volume is suddenly reduced 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 :

• A. Diminish to one-fourth of its initial value
• B. Diminish to one-eighth of its initial value
• C. Increase to eight times of its initial value
• D. Increase to four times of its initial value.

Answer 1

Answer: (C)

Increase to eight times of its initial value

Answer 2

$\text{Rat}\text{e}_{1}\text{ = k }\lbrack\text{NO}\rbrack^{2}\ \lbrack O_{2}\rbrack$

When volume is reduced to 1/2, concentration become two times,

Rate2 $= \mathrm { k } [ 2 \mathrm { NO } ] ^ { 2 } \left[ 2 \mathrm { O } _ { 2 } \right]$ $\frac{Rate_{1}}{Rate_{2}} = \frac{k\lbrack NO\rbrack^{2}\left\lbrack O_{2} \right\rbrack}{k\lbrack 2NO\rbrack^{2}\left\lbrack 2O_{2} \right\rbrack}$ or

$\frac{Rate_{1}}{Rate_{2}} = \frac{1}{8}$

$\therefore$ Rate2 = 8 Rate1.