At $${518^ \circ }C,$$  the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of $$363\,Torr,$$  was $$1.00\,Torr\,{S^{ - 1}}$$   when $$5\% $$  had reacted and $$0.5\,Torr\,{S^{ - 1}}$$   when $$33\% $$  had reacted. The order of the reaction is :

Solution :
$$\eqalign{ & C{H_3}CHO \to C{H_4} + CO \cr & {\text{Generally}}\,\,r \propto \left( {a - x} \right)m \cr & {r_1} = 1\,torr\,{s^{ - 1}},\,{\text{when}}\,{\text{5}}\% \,{\text{reacted}} \cr & {r_2} = 0.5\,torr\,{s^{ - 1}},\,{\text{when}}\,33\% \,{\text{reacted}} \cr & \left( {a - {x_1}} \right) = 0.95\left( {{\text{unreacted}}} \right) \cr & \left( {a - {x_2}} \right) = 0.67\left( {{\text{unreacted}}} \right) \cr & \frac{{{r_1}}}{{{r_2}}} = {\left[ {\frac{{\left( {a - {x_1}} \right)}}{{\left( {a - {x_2}} \right)}}} \right]^m};\frac{1}{{0.5}} = {\left( {\frac{{0.95}}{{0.67}}} \right)^m} \cr & 2 = {\left( {1.41} \right)^m} \Rightarrow 2 = {\left( {\sqrt 2 } \right)^m} \cr & \Rightarrow m = 2 \cr} $$