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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.

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$$r=k \times [NO_2]^2 \times [Cl_2]$$

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 $$r=1.21 \times 10^{-7}$$

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  • $\begingroup$ The question asked to determine the rate of formation of NOCl{R(NOCl)} ?IsR(NOCl) = r , corresponding to the stoichiometric coefficient of NOCl in this reaction equation : NO+(1/2)Cl2⟶NOCl or R(NOCl) = 2 r corresponding to the stoichiometric coefficient of NOCl in this reaction equation 2NO+Cl2⟶2NOCl $\endgroup$ – Adnan AL-Amleh Jan 26 '17 at 21:58
  • $\begingroup$ My bad, now I understand. You are right, it should be 2r for NOCl production. The answer provided is wrong. $\endgroup$ – mamun Jan 27 '17 at 0:27
  • $\begingroup$ If we multiply the reaction equation by 2 , does the rate of formation of NOCl change ?How much ?how and why ? $\endgroup$ – Adnan AL-Amleh Jan 27 '17 at 1:55
  • $\begingroup$ How the way of writing the chemical reaction -Duplication the reaction-affect on rate ? $\endgroup$ – Adnan AL-Amleh Mar 28 '17 at 13:20

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