1.

AT 518^(@)C , the rate of decomposition of a sample of gaseous acetaldehyde , initially at a pressure of 363 Torrr , was 1.00 Torr s^(-1) when 5% had reached and 0.5 Torr s^(-1) when 33% had reacted . The order of the reaction is

Answer»


Solution :RATE = K `("PRESSURE")^(N)`
`R_(1) = K(P_(1))^(n) , R_(2) = K(P_(2))^(n) implies (R_(1))/(R_(2)) = [(P_(1))/(P_(2))]^(n)`
`P_(1) = 363 - (363 xx 5)/(100) = 344. 85 , P_(2) = 363 - (363 xx 33)/(100) = 243.21`
`(1)/(0.5) = ((344.85)/(243.21))^(n) implies 2 = (sqrt2)^(n) , (n = 2)`.


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