Question

Trial [A] (M) [B] (M) [C] (M) Initial rate (M/s) 1 0.30 0.30 0.30 9.0×10−5 2 0.30 0.30 0.90 2.7×10−4 3 0.60 0.30 0.30 3.6×10−4 4 0.60 0.60 0.30 3.6×10−4 Given the data calculated in Parts A, B, C, and D, determine the initial rate for a reaction that starts with 0.85 M of reagent A and 0.90 M of reagents B and C?

Answer 1

From the given data we get the order with respect A as 2 , comparing 1 and 3 runs, as [A] doubles the rate become 4 times or 2^2. Thus we take order with respect to A is 2.

Similarly comparing runs 3 and 4 even after doubling [B] , no change in rate , shows order with respect to B is zero.

Then comparing runs 1 and 2 when [C] increases by 3 times the rate incresed by 3 times, means order with respect to C is one.

Thus the rate expression is written as rate = k [A]²[B]⁰[C]

To calculate the rate at different concentration we need to calculaate rate constant k.

By substituting any run values we can evaluate k value.

using run 1 rate = 9.0x10^-5 = k. (0.3)² (0.3)

Thus k = 1.0 x10^-2 / 3

Now Rate when [A] = 0.85 and [B] = [C]= 0.9 wll be

rate = 1.0 x10^-2x (0.85)²x (0.9) /3= 0.00216 M/s = 2.16x10^-3 M/s