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Consider the following reaction:

$$\ce{2 H2 (g) + 2 NO (g) -> N2 (g) + 2 H2O (g)}$$

The elementary steps of this reaction are:

$$ \begin{align} &1^\text{st}~\text{step:} &\ce{H2 (g) + 2 NO (g) &-> N2O (g) + H2O (g)} &&\text{(slow)} \\ &2^\text{nd}~\text{step:} &\ce{N2O (g) + H2 (g) &-> N2 (g) + H2O (g)} &&\text{(fast)} \\ \end{align} $$

The reaction is exothermic overall.

(a) What is the rate law for this reaction? Justify your answer.

$$\text{rate} = k[\ce{NO}]^2[\ce{H2}]$$

The 1st step of the reaction is the rate limiting step.

How did they know the order of the reaction with respect to $\ce{NO}$ was $2$, and $1$ for $\ce{H2}$? With other questions the orders are something they state in the question and are unrelated to the coefficients.

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You are correct, the reaction order cannot be deduced from the brutto-equation and is usually determined experimentally.

However, here you are presented with the elementary reactions and their rates. Obviously, the slowest step is the bottleneck and controls the overall kinetics, hence the total rate matches the rate of the slowest step. And as this is an elementary reaction, you just apply the law of mass action and use the stoichiometric coefficients for the rate equation.

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With other questions the orders are something they state in the question and are unrelated to the coefficients.

a- If the reaction is multi- step," the orders are may or may be not related to the coefficients" that the order and the rate equation of a given reaction cannot be reliably deduced from the stoichiometry and must be determined experimentally.

b- If the reaction is one step," the orders are related to the coefficients". In this case, the exponents in the rate law equation (or order of the reaction) are equal to the number of colliding molecules or to the stoichiometric coefficients of those reactants in the balanced equation that represents that collision. There is, of course, the possibility of bimolecular collisions, and the probability of termolecular collision is weak, especially for fast reactions. Four-molecule collisions are unlikely to occur in most cases.

How did they know the order of the reaction with respect to $\ce{NO}$ was $ 2$, and $1$ for$\ce{ H2}$?

The above-mentioned reaction mechanism has two steps. The first step is the slow step followed by a fast step. Since the total rate of the reaction is determined by the rate of the slow step, which is described as the Rate-Determining Step, so the rate of the reaction is equal to the rate of the reaction of the first slow step. $$\ce{\mathrm{r}= \mathrm{k}[{NO}]^2[{H_2}]}$$ $$\color{red}{\text{Another proposal for the mechanism of reaction}}$$ The first slow step of the above-mentioned mechanism in the question indicates that three molecules ( two molecules of $\ce{NO} $, one molecule of $\ce{H2}$) collide with each other at the same space and time (Termolecular Collisions), but the probability of termolecular collision is weak, so, there is another mechanism excludes termolecular collision as the following: \begin{array}{ }\ce{ \text{First step:}~~~2NO_\mathrm{(g)} <=>[k_\mathrm{1}][k_\mathrm{-1}]N2O2 &\quad\left (\mbox{Fast }\right)\\ \text{Second step:}~~~N2O2\mathrm{(g)}+H_2\mathrm{(g)} -> N2O\mathrm{(g)}+H2O \mathrm{(g)} &\quad\left(\mbox{Slow }\right)\\ \text{Third step:}~~~N2O\mathrm{(g)}+H2 \mathrm{(g)}-> N2\mathrm{(g)}+H2O \mathrm{(g)} &\quad\left(\mbox{Fast }\right) \\ }\end{array}

$$\color{red}{\text{ Describing the proposed reaction mechanism }}$$ The reaction mechanism consists of three steps where the first step is not the slow step, but the slow step precedes by a fast step followed by a fast step. In this case, where the slow step is preceded by a fast step, the first step reaches a state of equilibrium, beginning with the reverse of the reaction of the first step. This is because the first fast step produces a reactive intermediate material $\ce{ N2O4}$ at a fast rate, but the second slow step cannot consume these $\ce{ N2O4}$ at a similar rate, leading to the accumulation of $\ce{ N2O4}$, so, the probability of collisions of the $\ce{ N2O4}$ molecules increases, this leads to reverse the reaction in the first step, thus the reaction in the first step reaches at equilibrium. The forward and backward reactions are fast in the first step, but the reaction of the third step is fast in the forward direction

$$\color{red}{\text{ Deducing the order or rate equation from the reaction mechanism }}$$ Where $K_1$ is the rate constant of the forward reaction, as $K_{-1}$ is the rate constant for the backward reaction of the first step, and $K_2 $ is the rate constant for the reaction of the slow step. If the concentration of $\ce{ N2O2}$ which produced from the first step increased, the reaction rate increased in the opposite direction, leading to the arrival of the concentration intermediate material $\ce{ N2O4}$ to a fixed value. This is called Steady State, where the concentration of $\ce{N2O2}$ is constant, The rate of appearance of the intermediate material $\ce{N2O2}$ equal the rate of its disappearance, this material produced from the first step in the forward direction and consumed from the first step in the opposite direction, and in the second step also. $$\text{ Production rate of}~ \ce{N2O2} = \text{ Consumption rate of}~ \ce{N2O2}$$ $$k_\mathrm {1}\ce{[NO]^ 2}= k_\mathrm{-1}[\ce{N2O2}] + k_\mathrm {2} \ce{[NO] [H2]}$$ But $k_2 << k_{-1}$ because the second step is the slow step, so the term $k_2\ce{ [NO] [H2] }$ can be neglected at the far right of the last equation. $$k_1\ce{[NO]^2}= k_\mathrm{-1}\ce{ [N2O2] }$$ $$[\ce{N2O2}] =\frac{ k_1 [NO] ^ 2 }{ k_\mathrm{-1}}$$ The value of$ [\ce{N2O2}]$ from the last equation can now be compensated in the reaction rate equation which represents the slow step.

$$\text{Rate} = \ce{k_2[N2O2][H2] = \frac{k_1k_2}{k_\mathrm{-1}}\times{[NO]^2 [H2]} = k[[NO]^2[H2]}$$ In it we see that k expresses the value of the fractional $\frac{k_1k_2}{k_\mathrm{-1}}$

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