For a chemical reaction A to B , the rate of the reaction is 2 xx 10^(-3) mol dm^(-3) s^(-1) . When the initial concentration is 0.05 mol dm^(-3) . The rate of the same reaction is 1.6 xx 10^(-2) mol dm^(-3) s^(-1) when the initial concentration is 0.1 mol dm^(-3) . The order of the reaction is

For a chemical reaction A to B , the rate of the reaction is 2 xx 10^(

1.	For a chemical reaction A to B , the rate of the reaction is 2 xx 10^(-3) mol dm^(-3) s^(-1) . When the initial concentration is 0.05 mol dm^(-3) . The rate of the same reaction is 1.6 xx 10^(-2) mol dm^(-3) s^(-1) when the initial concentration is 0.1 mol dm^(-3) . The order of the reaction is
Answer» 0 3 1 2 Solution :Let the rate EQUATION for the REACTION be , rate = `k [A]^(n)` where , k = rate constant , n = order of reaction [A] = concentration of reactant given , `"rate"_(1) = 2 xx 10^(-3) "mol" dm^(-3) s^(-1)` `[A_(0)] = 0.05 ` mol `dm^(-3)` `"rate"_(2) = 1 . 6 xx 10^(-2)` mol `dm^(-3) s^(-1)` `(A_(0)) = 0.1` mol `dm^(-3)` `therefore 2 xx 10^(-3) = k [0.05]^(n) "".... (i)` `1.6 xx 10^(-2) = k[0.1]^(n) "".... (ii)` DIVIDE (i) by (ii) `implies (2 xx 10^-3)/(1.6 xx 10^(-2)) = ([0.05]^(n))/([0.1]^(n)) = ([0.05]^(n))/(2^(n)[0.05]^(n))` `implies (1)/(2^(n)) = (1)/(8) implies n = 3, "" therefore` Order of reaction = 3 .