# How to calculate the rate of formation of a product when given the rate law?

The following data are given for the reaction of $$\ce{NO} \text{ and } \ce{Cl2}$$:

$$\ce{2NO + Cl2 -> 2NOCl}$$

The reaction is second order in $$[\ce{NO}]$$ and first order in $$[\ce{Cl2}]$$, and the initial rate equals $$\pu{1.43E-6 mol L^-1 s^-1}$$ at the instant when $$[\ce{NO}]_0 = [\ce{Cl2}]_0 = \pu{0.25 mol L^-1}$$.

The problem then says:

Calculate the rate of formation of $$\ce{NOCl}$$ when $$[\ce{NO}] = [\ce{Cl2}] = \pu{0.11 mol L^-1}$$.

I thought it's just $$(2) × (\pu{9.152E-5})× [0.11]^2[0.11]$$, which is $$\pu{2.44E-7}$$, but the answer provided is half that $$(\pu{1.22E-7})$$, so I must have gotten the rate equation incorrect.

Is the rate of formation of $$\ce{NOCl}$$ equal to twice the rate of the reaction, as when I consider this reaction equation: $$\ce{2NO + Cl2 -> 2NOCl}$$?

Or is the rate of formation of $$\ce{NOCl}$$ directly equal to the rate of the reaction, as when I consider this reaction equation: $$\ce{NO + 1/2Cl2 -> NOCl}$$?

I'm really confused how to tell which of these two approaches to use.

$$r=k \times [NO_2]^2 \times [Cl_2]$$
From the initial condition information => $$k=\frac{r}{[NO_2]^2 \times [Cl_2]}$$ Plugging the numbers=> $$k=\frac{1.43 \times 10^{-6}}{0.25^2 \times 0.25}$$ So, $$k=9.152 \times 10^{-5}$$ For 0.11 condition => $$r=9.152 \times 10^{-5} \times 0.11^2 \times 0.11$$ or $$r=1.21 \times 10^{-7}$$