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Table of Contents
Rate of Reaction and Rate Law
For most equations you can write a rate law in the form of $r = k[X]^m[Y]^n$, where r is the rate of appearance of a product compound, $X$ and $Y$ being the reactant compounds, $k$ being the rate constant, and $m$ and $n$ being the order of the reactants.
This rate law equation allows you to make guesses on the reaction rate based off a $k$ value (given or calculated) alongside quantities of $X$ and $Y$.
Finding Orders
Take the example diagram here for the reaction $2NO_{(g)} + 2H_{2 (g)} -> N_{2 (g)} + 2H_2O_{(g)}$. To determine $m$ and $n$ for the rate law, you have to find pairs of trials that only modify 1 of the reactant compound's quantity, divide it, and compare the quotient to the reaction rate's quotient1) . Refer to the table below for the corresponding order.
| Muliplier | Order | Unit |
|---|---|---|
| No difference | 0th order | $M^1*s^{-1}$ |
| 2x | 1st order | $M^0*s^{-1}$ or $s^{-1}$ |
| 4x | 2nd order | $M^{-1}*s^{-1}$ |
| 8x | 3rd order | $M^{-2}*s^{-1}$ |
As $H_2$ has a 2:2 difference, $H_2$ is a 1st order reaction, consequently, because $NO$ is a 2:4 difference, it would be 2nd order. Thus, the rate law for this reaction is $r = k[H_2][NO]^2$.
Solving for $k$
We can use one of the table entries to solve for $k$ using the rate law we just created. The rate of production would be $r$ and the values for $NO$ and $H_2$ would be placed in their respective slots.
- $r = k[H_2][NO]^2$
- $1.8 \times 10^{-4} = k[0.001][0.006]^2$2)
- $\frac{1.8 \times 10^{-4}}{3.6 \times 10^{-8}} = k$
- $k = 5000$ $M^{-2}*s^{-1}$
The unit for each order is given to the right of the previously mentioned table, and the sum of the rate law's orders corresponding unit would be the $k$ value's unit.
Solving Rate of Production or $r$
Now that we have an equation and $k$ value, we can estimate rates of production of $N_2$ by plugging and chugging values into our new rate law. Take trial 5 for example:
- $r = k[H_2][NO]^2$
- $r = 5000[0.002][0.002]^2$
- $r = 4 \times 10^{-5}$ $\frac{\text{mol}}{L}*\text{min}$
Rate of Disappearance
Rate of Disappearance is similar to Production, after solving for the production of $N_2$, use stoichiometry to find the rate of disappearance. Looking back at our equation, there is a 2:1 ratio of $NO$ to $N_2$, so we'll use that.
$ 4 \times 10^{-5} * \frac{2NO}{1N_2} = -8 \times 10^{-5}$ $\frac{\text{mol}}{L}*\text{min}$3) which is our rate of disappearance of $NO$ for trial 5.
