At 518 o 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

Step-by-step solution

C H 3 C H O P i = 363 t o r r → C O + C H 4 Now we know that r ∝ P R n ...(i) where P R = reactant pressure n = order of reaction Now rate of reaction is 1.00 torr s - 1 , when reactant pressure is 363 - 363 × 5 100 t o r r = 344.85 t o r r , similarly rate of reaction is 0.5 torr s - 1 , when reactant pressure is 363 - 363 × 33 100 t o r r = 243.21 t o r r Therefore, applying equation (i) 1 0.5 = 344.85 243.21 n 2 = ( 1.418 ) n 2 1 n = 1.418 So n ≈ 2