Rate of Reaction and Rate Law: Quick Revision

A brief note on Rate of Reaction and Rate Law. Read it before a test. Try each question at the end before you open its answer.

Rate of a reaction

• Rate $= -\dfrac{\Delta[\text{reactant}]}{\Delta t} = +\dfrac{\Delta[\text{product}]}{\Delta t}$, in $\text{mol L}^{-1}\ \text{s}^{-1}$.
• For $a\text{A} + b\text{B} \rightarrow c\text{C} + d\text{D}$, divide each rate by its coefficient. The results are all the rate of the reaction.
• Average rate: the slope of a secant. Rate at one moment: the slope of the tangent. The initial rate is the rate at $t = 0$.
• Most reactions slow down as the reactants are used up.

Rate law and order

• Rate law: $\text{rate} = k[\text{A}]^x[\text{B}]^y$. The overall order is $x + y$.
• The orders come from experiment. They may be zero, whole numbers or fractions, and they need not match the coefficients.
• $k$ is the rate when every concentration is $1\ \text{mol/L}$. It depends on temperature but not on concentration.
• The unit of $k$ for order $n$ is $\text{L}^{n-1}\ \text{mol}^{1-n}\ \text{s}^{-1}$.
• Method of initial rates: change one concentration at a time. The rate changes by $f^x$ when $[\text{A}]$ changes by a factor $f$.

Molecularity and mechanism

• Molecularity: the number of particles that collide in one elementary step. It is 1, 2 or 3.
• Only an elementary step has a rate law that follows from its equation.
• A complex reaction goes in steps. The slowest step, the rate-determining step, sets the rate law.
• A mechanism must add up to the overall equation and give the measured rate law. A match does not prove it.

A worked example

The initial rate of a reaction of A and B is $3.0 \times 10^{-5}\ \text{mol L}^{-1}\ \text{s}^{-1}$ when $[\text{A}] = 0.10\ \text{mol/L}$ and $[\text{B}] = 0.10\ \text{mol/L}$. Doubling $[\text{A}]$ gives $1.2 \times 10^{-4}\ \text{mol L}^{-1}\ \text{s}^{-1}$. Doubling $[\text{B}]$ leaves the rate unchanged. Find the rate law and $k$.

1. Order in A. The rate becomes $\dfrac{1.2 \times 10^{-4}\ \text{mol L}^{-1}\ \text{s}^{-1}}{3.0 \times 10^{-5}\ \text{mol L}^{-1}\ \text{s}^{-1}} = 4$ times as large, so $x = 2$.
2. Order in B. The rate does not change, so $y = 0$.
3. The law. $\text{rate} = k[\text{A}]^2$. The overall order is 2.
4. The constant. $k = \dfrac{3.0 \times 10^{-5}\ \text{mol L}^{-1}\ \text{s}^{-1}}{(0.10\ \text{mol/L})^2} = 3.0 \times 10^{-3}\ \text{L mol}^{-1}\ \text{s}^{-1}$.

Practice with new numbers

A program to try

The program shows the unit of the rate constant for each overall order.

Mistakes to avoid

• Reading the orders from the balanced equation. They come from experiment.
• A negative rate. Put a minus sign in front of the change in a reactant.
• Changing two concentrations at once in the method of initial rates.
• A fractional molecularity. Molecularity is 1, 2 or 3. Only the order can be a fraction.
• Thinking the fastest step controls the rate. The slowest step does.

Check yourself

1. In $2\text{A} \rightarrow \text{B}$, A is used up at $0.080\ \text{mol L}^{-1}\ \text{s}^{-1}$. At what rate does B form?

2. What is the unit of $k$ for a first order reaction, and for a second order reaction?

3. Can an elementary step have a molecularity of two and a half?