The following reaction was studied at −10°C.

Question

2 NO(g) + Cl2(g) → 2 NOCl(g)
The following results were obtained where the rate of the reaction is given below.
[NO]o [Cl2]o INITIAL RATE
0.10 0.10 0.18
0.10 0.20 0.36
0.20 0.20 1.45

What is the rate law?
What is the value of the rate constant?

Answer 1

Don't you want to know how to do these yourself?
I see trial 1 and trial 2 has same concn for NO and double the cncn for Cl2. Note the k changes by 0.18 to 0.36 or double.
Cl2 doubled; rate doubled, so 2^x = 2 and x must be 1.

Then look at trial 2 and 3. Cl2 stays the same but NO doubles. The rate changes from 0.36 to 1.45 or (1.45/0.35 = 4.03) so 2^x = 4. x must be ?
Now take ANY of the three trials and write the rate constant expression.
rate = k[NO]^x[Cl2]^y and enter the rate for the trial you've chosen, plug in (NO), (Cl2), and x and y where x and y represent the values of x you have above.. Solve for k.