when to use exponents, or multiply by constants in the equation of the rate involving $k$ and concentrations?

Question

More specifically, I do not understand

rate of consuming $\ce{I}=2 k_2[\mathrm{I}]^2+2 k_3\left[\mathrm{H}_2\right][\mathrm{I}]^2$

Why is the concentration of $"\ce{I}"$ squared from $"2 \cdot k_3 \cdot \ce{[H_2][I]^2}"$ 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 [I] 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. $\mathrm{H}_{2(\mathrm{~g})}+2 \mathrm{I}_{(\mathrm{g})} \xrightarrow{k_3} 2 \mathrm{HI}_{(\mathrm{g})}$

Answer 1

To understand why we have the given rate of consumption of iodine ( $\mathrm{I}_2 $) for the reaction $\mathrm{H}_2 (\mathrm{g}) + \mathrm{I}_2 (\mathrm{g}) \rightarrow 2 \mathrm{HI} (\mathrm{g}) $), let's analyze the potential mechanisms and the rate laws that could lead to the given expression:

Reaction Mechanism and Rate Laws

In chemical kinetics, the overall rate law can often be explained by considering a series of elementary steps, each with its own rate law. Let's consider a possible mechanism for the given reaction:

Step 1: Initiation

$$ \mathrm{I}_2 \xrightarrow{k_1} 2\mathrm{I} \quad (\text{dissociation of iodine}) $$

Step 2: Propagation

$$ \mathrm{I} + \mathrm{H}_2 \xrightarrow{k_2} \mathrm{HI} + \mathrm{H} $$

$$ \mathrm{H} + \mathrm{I}_2 \xrightarrow{k_3} \mathrm{HI} + \mathrm{I} $$

Step 3: Termination

$$ 2\mathrm{I} \xrightarrow{k_4} \mathrm{I}_2 \quad (\text{recombination of iodine atoms}) $$

Derivation of Rate Law

To derive the rate law for the consumption of $\mathrm{I}_2 $, we need to consider the rates of the individual steps and how they contribute to the overall reaction.

1. Dissociation of $\mathrm{I}_2 $: $$ \frac{d[\mathrm{I}_2]}{dt} = -k_1[\mathrm{I}_2] $$

2. Rate of formation of $\mathrm{HI} $: The formation of $\mathrm{HI} $ occurs in the propagation steps. For each step involving $\mathrm{I} $ or $\mathrm{H} $:

   $$ \text{Step 2: } \mathrm{I} + \mathrm{H}_2 \rightarrow \mathrm{HI} + \mathrm{H} $$

   $$ \text{Rate} = k_2[\mathrm{I}][\mathrm{H}_2] $$

   $$ \text{Step 3: } \mathrm{H} + \mathrm{I}_2 \rightarrow \mathrm{HI} + \mathrm{I} $$

   $$ \text{Rate} = k_3[\mathrm{H}][\mathrm{I}_2] $$

   To link this to $\mathrm{I}_2 $ consumption, we consider the formation and consumption of intermediates ( $\mathrm{I} $ and $\mathrm{H} $):

3. Steady-State Approximation for $\mathrm{I} $ and $\mathrm{H} $: Assuming a steady-state for the intermediate $\mathrm{I} $:

   $$ \frac{d[\mathrm{I}]}{dt} = 0 = k_1[\mathrm{I}_2] - k_2[\mathrm{I}][\mathrm{H}_2] - k_4[\mathrm{I}]^2 $$

   Solving for $[\mathrm{I}] $:

   $$ [\mathrm{I}] = \sqrt{\frac{k_1[\mathrm{I}_2]}{k_4 + k_2[\mathrm{H}_2]}} $$

4. Substitution into Rate Law: Substitute $[\mathrm{I}] $ into the rate law expressions:

   $$ \frac{d[\mathrm{I}_2]}{dt} = -2k_2[\mathrm{I}]^2 - 2k_3[\mathrm{H}_2][\mathrm{I}]^2 $$

   Since $[\mathrm{I}] $ was derived as:

   $$ [\mathrm{I}] = \sqrt{\frac{k_1[\mathrm{I}_2]}{k_4 + k_2[\mathrm{H}_2]}} $$

   Squaring $[\mathrm{I}] $:

   $$ [\mathrm{I}]^2 = \frac{k_1[\mathrm{I}_2]}{k_4 + k_2[\mathrm{H}_2]} $$

   Substituting $[\mathrm{I}]^2 $ back into the rate expression:

   $$ \frac{d[\mathrm{I}_2]}{dt} = -2k_2 \left(\frac{k_1[\mathrm{I}_2]}{k_4 + k_2[\mathrm{H}_2]}\right) - 2k_3[\mathrm{H}_2] \left(\frac{k_1[\mathrm{I}_2]}{k_4 + k_2[\mathrm{H}_2]}\right) $$

Simplifying this expression yields the overall rate law. However, the given rate law might be an empirical rate law determined experimentally rather than derived from a proposed mechanism.

Given Rate Law

The given rate of consumption of $\mathrm{I}_2 $ is: $$ \frac{d[\mathrm{I}_2]}{dt} = -2 k_2[\mathrm{I}]^2 - 2 k_3[\mathrm{H}_2][\mathrm{I}]^2 $$

This suggests that both the iodine atoms' dimerization and their reaction with $\mathrm{H}_2 $ significantly impact the rate at which $\mathrm{I}_2 $ is consumed, consistent with the steps we considered.