1 and 2; C is same, B is same, A doubles and rate stays same; therefore, the reaction is zero order for A.
2 and 3. C is same. A doesn't matter since it is zero order. B doubles and rate doubles; therefore, the reaction is 1st order with in B.
1 and 4. A same. B same. C doubles while rate quadruples; therefore, the reaction is 2nd order in C.
rate = k(A)^o*B)^1(C)^2
Use any trial, substitute the values for concn of A, B, and C and solve for k.
2A + B2 + C --> A2B + BC
Run # Initial Rate (M-1s-1) (A)o (B)o
(C)o
1 1.6 x 106 1.2 M 1. x 10-2 M 1.0 M
2 1.6 x 106 2.4 M 1. x 10-2 M 1.0 M
3 3.2 x 106 1.2 M 2. x 10-2 M 1.0 M
4 6.4 x 106 1.2 M 1. x 10-2 M 2.0 M
What would be the rate-law expression of the reaction above?
1 answer
1 answer