R2.2 How fast? The rate of chemical change

If the concentration of a reactant decreases, $\Delta [\text{Reactant}]$ will be negative. However, rates are always reported as positive values, so we use the absolute value of the change.

Calculating the Rate of Reaction

For the reaction:
$4{\text{NH}}_3(g)+5{\text{O}}_2(g)\to 4\text{NO}(g)+6{\text{H}}_2\text{O}(g)$

Over 5 seconds, the concentration of NO increases from 0 to $6.0\times 10^{-3}$ mol dm${}^{-3}$.

$\text{Rate w.r.t. NO}=\frac{6.0\times 10^{-3}}{5}=1.2\times 10^{-3}{\text{ mol dm}}^{-3}{\text{ s}}^{-1}$

To find the rate with respect to O${}_2$: for every 4 mol NO produced, 5 mol O${}_2$ is consumed. Using the stoichiometric ratio:

${\text{Rate w.r.t. O}}_2=1.2\times 10^{-3}\times \frac{5}{4}=1.5\times 10^{-3}{\text{ mol dm}}^{-3}{\text{ s}}^{-1}$

Always check stoichiometry when comparing rates expressed in terms of different species , the rates are not necessarily equal unless the stoichiometric coefficients are the same.

Instantaneous Rate from a Graph

The concentration of H${}_2$ produced in a reaction is plotted over time. At $t=20$ s, a tangent to the curve passes through the points (10, 0.05) and (30, 0.20).

$\text{Slope}=\frac{0.20-0.05}{30-10}=\frac{0.15}{20}=0.0075{\text{ mol dm}}^{-3}{\text{ s}}^{-1}$

The instantaneous rate at $t=20$ s is 0.0075 mol dm${}^{-3}$ s${}^{-1}$.

To draw accurate tangent lines on exam graphs: use a ruler, ensure the line touches the curve at only one point, and does not cross it. Read off two well-separated coordinates to minimise error in your slope calculation.

Common mistake: Students often draw a chord (secant) between two points rather than a true tangent. A tangent just grazes the curve at a single point , it does not pass through two points on the curve.