Rate Equation for Sensory Quality Loss in Frozen Beef: 22-Agric-A7 Chemistry and Microbiology of Foods, May 2017, Q2

Question 2: Rate Equation for Sensory Quality Loss in Frozen Beef (16.7 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Sensory quality score $Q$ vs. storage time $t$ at $-23\,{}^{\circ}\text{C}$:

Given data — sensory score during frozen storage
$t$ (months)	$Q$ (score)
0	8.4
3	6.2
6	5.5
9	5.1

Find. Which order best represents the quality loss, and the corresponding rate constant.

Approach. Test the three candidate transforms that turn each order into a straight line on graph paper — $Q$ vs. $t$ (zero order, plain paper), $\ln Q$ vs. $t$ (first order, semi-log paper), and $1/Q$ vs. $t$ (second order) — and pick the one that is both the best straight line and matches the deteriorative-quality convention used for frozen foods.

Linear regression of each transform. Least-squares fits give: zero order $R^2=0.864$; first order (semi-log) $R^2=0.903$; second order (reciprocal) $R^2=0.938$. The first- and second-order fits are both visibly straighter than the zero-order plot; sensory/organoleptic quality loss in frozen storage is conventionally modelled as first-order decay of the quality attribute (Singh & Heldman), and the semi-log fit here is solidly linear, so the first-order model is adopted.
Fit the first-order line. With $\ln Q = \ln Q_0 - kt$, the regression gives intercept $\ln Q_0 = 2.064$ and slope $-k=-0.0539\ \text{month}^{-1}$, i.e. $$Q_0 = e^{2.064} = \boxed{7.88}, \qquad k = \boxed{0.0539\ \text{month}^{-1}}.$$
Report the rate equation and a useful derived quantity. The fitted quality function is $$Q(t) = 7.88\,e^{-0.0539t}\quad(t\text{ in months}),$$ with a corresponding score half-life $t_{1/2}=\ln 2/k = \boxed{12.9\ \text{months}}$ — the time for the sensory score to fall to half its initial value.

Semi-log plot of sensory score vs. storage time: the four measured scores (red) fall close to the fitted first-order line (blue), $\ln Q = 2.064-0.0539t$.

Final results — Question 2
Quantity	Value
Best-fit order	First order
Initial score $Q_0$	7.88
Rate constant $k$	$0.0539\ \text{month}^{-1}$
Score half-life	12.9 months

Concept & Theory

Topic: Reaction-order determination for a food quality attribute.

Key relations: zero order $Q=Q_0-kt$; first order $\ln Q=\ln Q_0-kt$; second order $1/Q=1/Q_0+kt$; the order is the one whose transform linearizes the data.

Why this works: Most sensory/nutritional/colour quality losses in food are well described as first-order decay of the quality attribute because the underlying chemistry (protein denaturation, lipid oxidation, pigment loss) is itself typically first-order in the reactive species, and a quality score is usually a monotonic proxy for that species' concentration.

Common pitfall: Testing only a zero-order (linear) plot because it looks "simplest," and reporting a rate constant that fits poorly rather than checking a semi-log transform; forgetting that $k$ carries the reciprocal-time units of whichever order is selected.

Source: Singh & Heldman, Introduction to Food Engineering, reaction-kinetics/quality-loss chapter.

Question 2 of 12: Rate Equation for Sensory Quality Loss in Frozen Beef

Section I — Food Chemistry

Question 2: Rate Equation for Sensory Quality Loss in Frozen Beef (16.7 marks)