**Step 1: Draw a tangent to the curve at t = 45 seconds.**
A straight line should be drawn that touches the curve at the 45 s point and has the same slope as the curve at that point.
**Step 2: Form a large triangle using the tangent to find the gradient.**
For example, the tangent might pass through (20, 0.004) and (80, 0.012).
(Note: any large triangle using the drawn tangent is acceptable).
**Step 3: Calculate the gradient of the tangent (change in y / change in x).**
Gradient = (0.012 - 0.004) / (80 - 20)
Gradient = 0.008 / 60
Gradient = 0.0001333... mol/s
**Step 4: Convert the answer to standard form.**
Rate = 1.33 x 10⁻⁴ mol/s
(Acceptable range for rate is typically 1.2 x 10⁻⁴ to 1.5 x 10⁻⁴ mol/s depending on the tangent drawn).
The rate of reaction at a particular instant in time is represented by the gradient (steepness) of the curve at that point. Since the graph is a curve, the gradient is constantly changing. To find the gradient at a specific point (t = 45 s), we must draw a tangent.
1. A tangent is a straight line that touches the curve at a single point without crossing it, and which has the same slope as the curve at that point.
2. Once the tangent is drawn, we can calculate its gradient, which is equal to the rate of reaction. The gradient is calculated as the change in the y-axis value divided by the change in the x-axis value (Δy/Δx).
3. We pick two points on the tangent line that are far apart to minimise reading errors. For example, using the points (20 s, 0.004 mol) and (80 s, 0.012 mol):
Δy = 0.012 - 0.004 = 0.008 mol
Δx = 80 - 20 = 60 s
Rate = Δy / Δx = 0.008 / 60 = 0.000133... mol/s.
4. To express this in standard form, we move the decimal point four places to the right, so the exponent is -4.
Rate = 1.33 x 10⁻⁴ mol/s.
Expert