Rate Calculations

Worked Example: Volume of Product

In an experiment, catalase breaks down hydrogen peroxide. The oxygen gas produced is collected using a gas syringe. The reaction produces $45\text{ cm}^3$ of oxygen gas in $5\text{ minutes}$. Calculate the rate of reaction.

Step 1: Convert time into standard units (seconds).

• $5\text{ minutes} \times 60 = 300\text{ s}$

Step 2: Substitute the experimental values into the formula.

• $\text{Rate} = 45 \div 300$

Step 3: Calculate the final answer with units.

• $\text{Rate} = 0.15\text{ cm}^3/\text{s}$

(Note: For Higher Tier papers, be prepared to write units using index notation, such as $0.15\text{ cm}^3\text{ s}^{-1}$).

Calculating Relative Rate Using 1/Time

You can easily measure the volume of a gas produced in a syringe, but try measuring the exact volume of starch disappearing on a spotting tile. For experiments with a fixed endpoint rather than a measurable volume, we calculate a "relative rate".

This is central to investigating the effect of pH on amylase. You record the time taken for starch to be completely broken down, indicated by the point where added iodine does NOT turn blue-black and stays orange/brown. The formula used is:

$$\text{Rate} = \frac{1}{\text{time}}$$

Worked Example: Fixed Endpoint Rate

Amylase breaks down starch at pH 6. It takes $50\text{ seconds}$ for the iodine to stay orange/brown.

Step 1: Write down the formula.

• $\text{Rate} = 1 \div \text{time}$

Step 2: Substitute the experimental value.

• $\text{Rate} = 1 \div 50$

Step 3: Calculate the final answer with standard units.

• $\text{Rate} = 0.02\text{ s}^{-1}$

Because calculating $1 \div \text{time}$ often results in very small decimals that are difficult to plot on a graph, examiners frequently suggest using a multiplier. You can use $\text{Rate} = \frac{1000}{\text{time}}$ to produce easy-to-graph whole numbers (e.g., $$ arbitrary units).

Using Tangents to Measure Rate on a Graph

Why do graphs of enzyme reactions curve instead of going up in a straight line? As the substrate is used up, the rate constantly drops. Eventually, the line forms a flat plateau, meaning the reaction has finished and the rate is zero.

To find the instantaneous rate at a specific point on the curve, you must draw a tangent. This is a straight line that touches the curve at a single point, but does NOT cross through it. You then calculate the gradient (steepness) of this line.

Worked Example: Tangent Gradient

Calculate the rate of reaction at $20\text{ seconds}$ from a curved volume-time graph.

Step 1: Draw the tangent.

• Find $20\text{ seconds}$ on the x-axis and move up to the curve.
• Place a ruler on the "outside" of the curve so it just touches the point (the "no daylight" rule).
• Draw a long straight line.

Step 2: Pick two easy-to-read points on your tangent line.

• Point 1: $(0\text{ s}, 10\text{ cm}^3)$
• Point 2: $(40\text{ s}, 30\text{ cm}^3)$

Step 3: Substitute into the gradient formula (Change in y $\div$ Change in x).

• $\text{Change in y} = 30 - 10 = 20\text{ cm}^3$
• $\text{Change in x} = 40 - 0 = 40\text{ s}$

Step 4: Calculate the final answer with units.

• $\text{Rate} = 0.5\text{ cm}^3/\text{s}$