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  4. How will the rate of reaction, 2NO+O2 xrightarrow[]2NO2 change, if the volume of the reaction vessel is doubled?

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Q. How will the rate of reaction, $ 2NO+{{O}_{2}}\xrightarrow[{}]{{}}2N{{O}_{2}} $ change, if the volume of the reaction vessel is doubled?

MGIMS WardhaMGIMS Wardha 2015

Solution:

When volume is doubled, concentration are halved. Hence, new rate $ =k{{\left( \frac{a}{2} \right)}^{2}}\left( \frac{b}{2} \right)=\frac{1}{8}K{{a}^{2}}b $ i.e. it is $ \frac{1}{8} $ th of the initial value.