Gradients as Rate

• Volume graphs: Units are usually $\text{cm}^3/\text{s}$ or $\text{cm}^3\text{ s}^{-1}$.
• Mass graphs: Units are usually $\text{g/s}$ or $\text{g s}^{-1}$.

Worked Example

A student investigates the reaction between magnesium and hydrochloric acid by measuring the volume of hydrogen gas produced over time. Calculate the rate of reaction at exactly 25 seconds.

Step 1: Locate 25 seconds on the x-axis, find the corresponding point on the curve, and draw a tangent.

Step 2: Construct a large right-angled triangle on the tangent and pick two easy-to-read coordinates.

• Point 1: $(5\text{ s}, 12\text{ cm}^3)$
• Point 2: $(45\text{ s}, 60\text{ cm}^3)$

Step 3: Calculate the change in y ($\Delta y$) and the change in x ($\Delta x$).

• $\Delta y = 60\text{ cm}^3 - 12\text{ cm}^3 = 48\text{ cm}^3$
• $\Delta x = 45\text{ s} - 5\text{ s} = 40\text{ s}$

Step 4: Substitute the values into the gradient formula.

• $\text{Rate} = \frac{48\text{ cm}^3}{40\text{ s}}$

Step 5: Calculate the final answer and include appropriate units.

• $\text{Rate} = 1.20\text{ cm}^3/\text{s}$

Initial Rate and Common Pitfalls

Sometimes, an exam question will ask for the initial rate of a reaction. This is simply the instantaneous rate at the very start of the reaction ($t = 0\text{ s}$). To find this, draw your tangent touching the curve exactly at the origin $(0,0)$ and calculate its gradient.

When calculating the rate at a specific time, you must avoid interpolation. Interpolation means taking a direct reading from a point on the curve (just reading the y-value and dividing it by the x-value). This is mathematically incorrect for finding the instantaneous rate on a curve because it does not use the slope. You must physically draw the tangent to calculate the true gradient.