[sd79812]

More specifically, I do not understand

>rate of consuming $\ce{I}=2 k_2[\mathrm{I}]^2+2 k_3\left[\mathrm{H}_2\right][\mathrm{I}]^2$
aWhy is the concentration of $"\ce{I}"$ squared from $"2 \cdot k_3 \cdot \ce{[H_2]^2}"$ when the reactant side's $"2\ce{I}"$ doesn't have a subscript of two on the "I", and the product side doesn't have a subscript of 2 on the "I" in "2HI"?

Are the exponents or the two multiplied by the $k_3$ or $k_2$ from the stoichiometric coefficients or from the subscripts?
Does the exponent on have to agree with the constant you multiple $k_2$ or $k_3$ by?

I'd like to learn how to do this on a general multistep mechanism.

>$$
\ce{H_{2 (g)} + I_{2 (g)} -> 2 HI_{(g)}}
$$

Elementary Reactions we propose as the mechanism:
These radicals are active, and they react with $\mathrm{H}_2$ to produce the products. Thus we propose the three-step mechanism:

>i. $\mathrm{I}_2(\mathrm{~g}) \xrightarrow{k_1} 2 \mathrm{I}_{(\mathrm{g})}$

>ii. $2 \mathrm{I}_{(\mathrm{g})} \xrightarrow{k_2} \mathrm{I}_{2(\mathrm{~g})}$

>iii. $\mathrm{H}_{2(\mathrm{~g})}+2 \mathrm{I}_{(\mathrm{g})} \xrightarrow{k_3} 2 \mathrm{HI}_{(\mathrm{g})}$

[Hunter2]

Sorry it's not readable.

[Babcock_Hall]

I second what Hunter2 said.  The exponents in rate laws may or may not be equal to the stoichiometric coefficients in the balanced equation in a multistep reaction.  If a reaction mechnism consists of a single step, that is another matter.

[sd79812]

More specifically, I do not understand

>rate of consuming $$\ce{I}=2 k_2[\mathrm{I}]^2+2 k_3\left[\mathrm{H}_2\right][\mathrm{I}]^2$$
aWhy is the concentration of $$"\ce{I}"$$ squared from $$"2 \cdot k_3 \cdot \ce{[H_2]^2}"$$ when the reactant side's $$"2\ce{I}"$$ doesn't have a subscript of two on the "I", and the product side doesn't have a subscript of 2 on the "I" in "2HI"?

Are the exponents or the two multiplied by the $$k_3$$ or $$k_2$$ from the stoichiometric coefficients or from the subscripts?
Does the exponent on have to agree with the constant you multiple $$k_2$$ or $$k_3$$ by?

I'd like to learn how to do this on a general multistep mechanism.

>$$
\ce{H_{2 (g)} + I_{2 (g)} -> 2 HI_{(g)}}
$$

Elementary Reactions we propose as the mechanism:
These radicals are active, and they react with $$\mathrm{H}_2$$ to produce the products. Thus we propose the three-step mechanism:

>i. $$\mathrm{I}_2(\mathrm{~g}) \xrightarrow{k_1} 2 \mathrm{I}_{(\mathrm{g})}$$

>ii. $$2 \mathrm{I}_{(\mathrm{g})} \xrightarrow{k_2} \mathrm{I}_{2(\mathrm{~g})}$$

>iii. $$\mathrm{H}_{2(\mathrm{~g})}+2 \mathrm{I}_{(\mathrm{g})} \xrightarrow{k_3} 2 \mathrm{HI}_{(\mathrm{g})}$$