Integrated Rate Equation — Zero Order Reactions

Zero Order Reactions

A zero-order reaction is one whose rate is independent of the concentration of the reactant — it stays constant as the reaction proceeds. For $R \rightarrow P$:

$$\text{Rate} = -\frac{d[R]}{dt} = k[R]^0 = k$$

Integrating from $[R]_0$ (at $t=0$) to $[R]$ (at time $t$):

$$\boxed{[R] = [R]_0 - kt}$$

This is the equation of a straight line ($y = c + mx$ form) when $[R]$ is plotted against $t$: the line starts at $[R]_0$ and has slope $-k$.

The units of $k$ for a zero-order reaction are mol L$^{-1}$ s$^{-1}$ (same as rate, since rate $= k$).

Graph and Half-Life of a Zero-Order Reaction

A plot of $[R]$ against time is a descending straight line:

$$\text{slope} = -k, \qquad \text{intercept} = [R]_0$$

Half-life ($t_{1/2}$) is the time for $[R]$ to fall to half of $[R]_0$. Putting $[R] = [R]_0/2$:

$$\frac{[R]_0}{2} = [R]_0 - k\,t_{1/2} \quad\Rightarrow\quad \boxed{t_{1/2} = \frac{[R]_0}{2k}}$$

So for a zero-order reaction, half-life is directly proportional to the initial concentration.

Real examples: the decomposition of gaseous ammonia on a hot platinum surface, and the decomposition of $\text{HI}$ on a gold surface, are zero order — the metal surface is saturated, so adding more reactant doesn't speed things up. Many enzyme-catalysed reactions are zero order at high substrate concentration.

Solved Examples

Example 1: Rate constant of a zero-order reaction

In a zero-order reaction, $[R]$ falls from 0.10 M to 0.075 M in 50 s. Find $k$.

Solution: $k = \dfrac{[R]_0 - [R]}{t} = \dfrac{0.10 - 0.075}{50} = \dfrac{0.025}{50} = 5\times10^{-4}$ mol L$^{-1}$ s$^{-1}$.

Example 2: Concentration after a given time

For a zero-order reaction with $k = 2\times10^{-3}$ mol L$^{-1}$ s$^{-1}$ and $[R]_0 = 0.50$ M, find $[R]$ after 100 s.

Solution: $[R] = [R]_0 - kt = 0.50 - (2\times10^{-3})(100) = 0.50 - 0.20 = 0.30$ M.

Example 3: Half-life of a zero-order reaction

A zero-order reaction has $k = 0.02$ mol L$^{-1}$ s$^{-1}$ and $[R]_0 = 0.10$ M. Find its half-life.

Solution: $t_{1/2} = \dfrac{[R]_0}{2k} = \dfrac{0.10}{2\times0.02} = \dfrac{0.10}{0.04} = 2.5$ s.

Example 4: Time for complete reaction

For the reaction in Example 3, how long until the reactant is completely consumed ($[R] = 0$)?

Solution: $[R] = [R]_0 - kt = 0 \Rightarrow t = \dfrac{[R]_0}{k} = \dfrac{0.10}{0.02} = 5$ s. (Note this is exactly $2\,t_{1/2}$ for zero order.)

Example 5: Identify zero order from units

A reaction's rate constant is $1.5\times10^{-3}$ mol L$^{-1}$ s$^{-1}$. What is its order, and how do you know?

Solution: Units mol L$^{-1}$ s$^{-1}$ are the units of $k$ for a zero-order reaction (they equal the units of rate). So the reaction is zero order.

Example 6: Slope of the graph

For a zero-order reaction, a plot of $[R]$ vs $t$ has slope $-4\times10^{-4}$ mol L$^{-1}$ s$^{-1}$. What is $k$?

Solution: slope $= -k$, so $k = 4\times10^{-4}$ mol L$^{-1}$ s$^{-1}$.

Example 7: Half-life proportional to concentration

If the initial concentration of a zero-order reaction is doubled, how does its half-life change?

Solution: $t_{1/2} = \dfrac{[R]_0}{2k}$, so $t_{1/2} \propto [R]_0$. Doubling $[R]_0$ doubles the half-life.

Example 8: Concentration from data

A zero-order reaction ($k = 1\times10^{-2}$ mol L$^{-1}$ min$^{-1}$) starts at 0.25 M. What is $[R]$ after 10 minutes?

Solution: $[R] = 0.25 - (1\times10^{-2})(10) = 0.25 - 0.10 = 0.15$ M.

Example 9: Finding time to reach a concentration

A zero-order reaction has $[R]_0 = 0.80$ M and $k = 0.05$ mol L$^{-1}$ s$^{-1}$. How long to reach $[R] = 0.50$ M?

Solution: $t = \dfrac{[R]_0 - [R]}{k} = \dfrac{0.80 - 0.50}{0.05} = \dfrac{0.30}{0.05} = 6$ s.

Example 10: Why surface reactions are zero order

Explain why the decomposition of ammonia on a hot platinum surface is zero order.

Solution: The platinum surface becomes fully saturated with ammonia molecules. Once saturated, adding more ammonia cannot increase the number reacting, so the rate stays constant — independent of ammonia concentration — i.e. zero order.