Overall Order of Reaction
The Building Blocks: Rate Laws and Rate Constants
Before we can add up exponents, we need to understand where they come from. Every reaction has a rate law, an equation that links the reaction speed to the concentrations of the reactants. It has a very specific form:
Let's break down each part:
- Rate: The speed of the reaction, often measured in $ \text{mol L}^{-1} \text{s}^{-1} $ (moles per liter per second).
- $ k $: The rate constant. This is a special number that is constant for a given reaction at a specific temperature. It's like the reaction's unique "speed factor."
- $ [A] $ and $ [B] $: The concentrations of reactants A and B, usually in moles per liter ($ \text{mol/L} $ or $ M $).
- $ m $ and $ n $: The reaction orders with respect to A and B. These are not the coefficients from the balanced chemical equation! They must be determined experimentally. They are usually small whole numbers (0, 1, 2) but can sometimes be fractions.
How to Find the Overall Order: Simple Examples
The overall order is found by adding the individual orders ($ m + n + ... $). Let's look at some common cases.
| Rate Law | Order in A ($ m $) | Order in B ($ n $) | Overall Order ($ m+n $) | What It Means |
|---|---|---|---|---|
| $ \text{Rate} = k $ | 0 | - | 0 | The rate is constant. Changing [A] has no effect on speed. |
| $ \text{Rate} = k[A] $ | 1 | - | 1 (First Order) | If you double [A], the rate also doubles. |
| $ \text{Rate} = k[A]^2 $ | 2 | - | 2 (Second Order) | If you double [A], the rate quadruples (2^2 = 4 times faster). |
| $ \text{Rate} = k[A][B] $ | 1 | 1 | 2 (Second Order) | First order in A, first order in B. Doubling either one alone doubles the rate. Doubling both makes it 4 times faster. |
| $ \text{Rate} = k[A]^{1/2}[B] $ | 1/2 | 1 | 1.5 (Three-halves Order) | Orders can be fractions! This is less common but possible. |
From Baking Cookies to Launching Rockets: Why Order Matters
The overall order of a reaction isn't just a number; it has real, practical consequences. It tells us exactly how sensitive the reaction rate is to how much "stuff" we start with.
Imagine you are following a cookie recipe. The recipe says the baking time depends on how much sugar you add in a special way: "If you double the sugar, the baking time is cut to one-fourth." That's a specific relationship! If you know this, you can predict exactly what happens if you triple the sugar or use only half. In chemistry, the overall order gives us this precise predictive power.
Let's see this with a concrete, second-order reaction example. Suppose we have a reaction with the rate law: $ \text{Rate} = k[NO_2]^2 $ This means the reaction is second order overall (and second order with respect to $ NO_2 $).
- If the initial concentration of $ NO_2 $ is $ 1.0 \text{ mol/L} $, let's say the rate is $ 0.01 \text{ mol L}^{-1}\text{s}^{-1} $.
- What happens if we increase the concentration to $ 2.0 \text{ mol/L} $ (double it)? The new rate = $ k \times (2.0)^2 = k \times 4.0 $. Since $ k $ is constant, the rate becomes 4 times the original rate: $ 0.04 \text{ mol L}^{-1}\text{s}^{-1} $.
- If we decrease it to $ 0.5 \text{ mol/L} $ (half it), The new rate = $ k \times (0.5)^2 = k \times 0.25 $. The rate becomes one-fourth of the original: $ 0.0025 \text{ mol L}^{-1}\text{s}^{-1} $.
This concept is vital in chemical engineering and manufacturing. For example, in designing a factory to make fertilizer, engineers need to know how the production speed will change if they use more concentrated raw materials. If the reaction is first-order, doubling concentration doubles output. If it's second-order, they get a much bigger boost! This helps them design safer, more efficient, and cost-effective processes.
Important Questions
No, absolutely not. This is a very common point of confusion. The coefficients from the balanced equation (like the 2 in $ 2H_2 + O_2 \rightarrow 2H_2O $) tell us the ratios in which molecules react. The orders in the rate law ($ m $ and $ n $) are determined experimentally and have no necessary relation to those coefficients. For the reaction between hydrogen and oxygen, the rate law might be something like $ \text{Rate} = k[H_2][O_2] $, making the overall order 2, not 3 (which would be 2+1 from the coefficients).
Yes. A zero-order reaction has a rate law like $ \text{Rate} = k $. The rate is completely independent of the concentration of the reactant(s). A real-world example is the decomposition of nitrous oxide ($ N_2O $) on a hot platinum surface: $ 2N_2O(g) \rightarrow 2N_2(g) + O_2(g) $. The platinum surface gets saturated with gas molecules, so adding more gas doesn't speed up the reaction. The reaction chugs along at a constant rate until one of the reactants is completely used up.
They use the Method of Initial Rates. They run the reaction multiple times, each time starting with different, known concentrations of the reactants. They carefully measure the initial rate at the very beginning of each run. By comparing how the initial rate changes when only one reactant's concentration is changed (while keeping the others constant), they can deduce the order for that reactant. For example, if doubling [A] doubles the rate, the order for A is 1. If doubling [A] quadruples the rate, the order for A is 2. Once all individual orders are known, they are added to get the overall order.
Footnote
1 Concentration: The amount of a substance (solute) present in a given volume of solution. Common unit: moles per liter ($ \text{mol L}^{-1} $ or M).
2 Rate Equation / Rate Law: An equation that links the reaction rate to the concentrations of reactants, each raised to a specific power (the order).
3 Chemical Kinetics: The branch of chemistry that studies the speeds (rates) of chemical reactions and the mechanisms by which they occur.
4 Rate Constant ($ k $): The proportionality constant in the rate law. Its value is specific to a particular reaction at a particular temperature.
5 Method of Initial Rates: An experimental technique to determine the orders of a reaction by measuring how the initial rate changes with different starting concentrations.