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The initial rate of reaction A + B to C, was measured for several concentration of A and B. The decomposition made are recorded in the following table. Using the data in the above table, determine a) the rate law for the reason b) the magnitude of the rate constant. |
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Answer» Rate =`k[A]^(p)[B]^(q)` Comparing experiments (1) and (2), `("Rate")_(1)=k[0.1]^(p)[0.1]^(q)=4.0 xx 10^(-5)`…………(i) `("Rate")_(2) = k[0.1]^(p)[0.2]^(q)=4.0 xx 10^(-5)`…………….(ii) DIVIDING EQN. (ii) by (i), `("Rate")_(2)/("Rate")_(1) = (k[0.1]^(p)[0.2]^(q))/(k[0.1]^(p)[0.1]^(q))/(k[0.1]^(p)[0.1]^(q)) = (4.0 xx 10^(-5))/(4.0 xx 10^(-5))` =1 `[2]^(q) = [2]^(0)` or q=0 Comparing experiments (1) and (3), `("Rate")_(1) = (k[0.1]^(p)[0.1]^(q))/([k[0.1]^(p)[0.1]^(q))) = (16.0 xx 10^(-5))/(4.0 xx 10^(-5)) = 4=[2]^(p)=[2]^(q)` or p=2 Order w.r. to A=2, Order w.r. to B=0 Rate law for the reaction `=k[A]^(2)` VALUE of rate constant from expt:1. `(4.0 xx 10^(-5) mol L^(-1)s^(-1)) =k[A] = k(0.1 mol L^(-1))^(2)`. `k=(4.0 xx 10^(-5) mol L^(-1)s^(-1))/(0.1 mol L^(-1))^(2) = 4.0 xx 10^(-3) L mol^(-1)s^(-1)` |
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