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In a reaction between A and B, the initial rate of reaction was measured for different initial concentration of A and B as given ahead: what is the order of reaction with respect to A and B? |
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Answer» Solution :The rate law equation may be expressed as: Rate = `k[A]^(p)[B]^(q)` Comparing experiments 1 and 2. `("Rate")_(1) = k[0.2]^(p)[0.3]^(q)= 5.07 XX 10^(-5)`..........(i) `("Rate")_(2) = k[0.2]^(p)[0.1]^(q) = 5.07 xx 10^(-5)`...........(ii) Dividing EQN. (i) by eqn. (ii), `(Rate)_(1)/(Rate)_(2) = (k[0.2]^(p)[0.3]^(q))/(k[0.2]^(p)[0.1]^(q)=(5.07 xx 10^(-5))/(5.07 xx 10^(-5)` `[3]^(q) = [1]^(0)` or q=0 By comparing experiments 2 and 3. `("Rate")_(2)= k[0.2]^(p)[0.1]^(q) = 5.07 xx 10^(-5)`...........(i) `("Rate")_(3) = k[0.4]^(P)[0.05]^(q)= 7.56 xx 10^(-5)`..............(iv) Dividing eqn. (iv) by eqn. (iii), `("Rate")_(3)/("Rate")_(2) = (k[0.4]^(p)[0.05]^(q))/(k[0.2]^(p)[0.1]^(q))= (7.56 xx 10^(-5))/(5.07 xx 10^(-5))` `[2]^(p) = [1.49]=[2]^(3//2)` or p `=3//2` Order with RESPECT to A = `3//2` Order with respect to B = 0 |
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