krit.club logo

Measurement and Data Processing - Graphical techniques

Grade 12IBChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

The independent variable is plotted on the horizontal axis (xx-axis) and the dependent variable is plotted on the vertical axis (yy-axis).

•

A line of best fit (trend line) should be drawn to represent the mathematical relationship; it does not necessarily pass through all data points due to random errors.

•

A direct proportionality exists if the graph is a straight line passing through the origin (0,0)(0,0), expressed as y=kxy = kx.

•

The gradient (mm) of a linear graph y=mx+cy = mx + c represents the rate of change of yy with respect to xx.

•

Error bars are used to represent the absolute uncertainty of the data points in both the xx and yy directions.

•

The uncertainty in the gradient can be estimated by drawing the lines of maximum and minimum slope that pass through all error bars.

•

Interpolation is the process of estimating a value within the range of measured data points, while extrapolation is estimating a value outside the range (which is less reliable).

•

Non-linear relationships can often be linearized; for example, plotting PV=nRTPV = nRT as PP vs 1V\frac{1}{V} yields a straight line.

📐Formulae

m=ΔyΔx=y2−y1x2−x1m = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}

y=mx+cy = mx + c

Δm=mmax−mmin2\Delta m = \frac{m_{max} - m_{min}}{2}

Δc=cmax−cmin2\Delta c = \frac{c_{max} - c_{min}}{2}

ln⁡k=−EaR(1T)+ln⁡A\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A

💡Examples

Problem 1:

A student measures the mass (mm) and volume (VV) of several samples of an unknown liquid to determine its density (ρ\rho). The line of best fit for the plot of mm (y-axis) against VV (x-axis) has a gradient of 0.850.85 and a y-intercept of 0.020.02. If mass is in gg and volume is in cm3cm^3, determine the density and identify the likely source of the non-zero intercept.

Solution:

The density ρ=0.85 g cm−3\rho = 0.85 \text{ g cm}^{-3}. The y-intercept c=0.02 gc = 0.02 \text{ g} is close to zero, but its presence suggests a systematic error, such as failing to tare the balance used to measure the liquid's mass.

Explanation:

In the equation m=ρV+cm = \rho V + c, the gradient mm represents the density ρ\rho. In an ideal scenario, mass is directly proportional to volume (c=0c=0), so a non-zero intercept indicates a constant offset in measurement.

Problem 2:

Given a graph of ln⁡k\ln k against 1T\frac{1}{T} for a chemical reaction, the gradient of the line is calculated to be −12000 K-12000 \text{ K}. Calculate the activation energy (EaE_a) of the reaction using the gas constant R=8.31 J K−1 mol−1R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1}.

Solution:

m=−EaRm = -\frac{E_a}{R} −12000=−Ea8.31-12000 = -\frac{E_a}{8.31} Ea=12000×8.31=99720 J mol−1≈99.7 kJ mol−1E_a = 12000 \times 8.31 = 99720 \text{ J mol}^{-1} \approx 99.7 \text{ kJ mol}^{-1}

Explanation:

The Arrhenius equation is linearized as ln⁡k=−EaR(1T)+ln⁡A\ln k = -\frac{E_a}{R}(\frac{1}{T}) + \ln A. By comparing this to y=mx+cy = mx + c, we see the gradient mm is equal to −EaR-\frac{E_a}{R}.

Problem 3:

How is the uncertainty in the gradient calculated if the maximum possible gradient is 1.251.25 and the minimum possible gradient is 1.151.15?

Solution:

Uncertainty (Δm)=1.25−1.152=0.102=0.05\text{Uncertainty } (\Delta m) = \frac{1.25 - 1.15}{2} = \frac{0.10}{2} = 0.05 The gradient would be reported as 1.20±0.051.20 \pm 0.05.

Explanation:

To find the uncertainty in the slope, IB students draw the 'best fit' line, the 'steepest' line (max gradient), and the 'shallowest' line (min gradient) that still pass through the error bars. The uncertainty is half the difference between the extremes.

Graphical techniques - Revision Notes & Key Formulas | IB Grade 12 Chemistry