Question

Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimental data provided.

Q + X yields products


Trial [Q] [X] Rate
1 0.12 M 0.10 M 1.5 × 10-3 M/min
2 0.24 M 0.10 M 3.0 × 10-3 M/min
3 0.12 M 0.20 M 12.0 × 10-3 M/min

Answer 1

rate1 = k1*(Q1)^a*(X1)^b where a and be are the exponents for the orders and (Q) and (X1) are the concentrations for rate 1.
rate2 = k2*(Q2)^a*(X2)^b

Then
rate1......(k1*(Q1)^a*(X)^b
------------------------
rate2......(k2)*(Q2)^a*(X)^b

Pick trial 1 and 2 since (X) is the same.
1.5E-3.......(0.12)^a*(0.1)^b
-----------------------------
3.0E-3.......(0.24)^a*(0.1)^b

Since (0.1)^b in the numerator and denominator cancel,(rember k1 and k2 are equal)you have only a as the unknown and you can solve for that.
1/2 = (1/2)^a
so a must be 1.
Determine b the same way by using trials 1 and 3.
After you know a and b, go back to ANY trial, plug in (Q)and (X) and a and b and rate and solve for k.

Answer 2

0.012

Answer 3

joe beans