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Cane sugar (sucrose) is hydrolysed in dilute hydrochloric acid at 298 K: C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Cane sugar (sucrose) is hydrolysed in dilute hydrochloric acid at $298\ \text{K}$ :

\displaystyle \text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6\ (\text{glucose}) + \text{C}_6\text{H}_{12}\text{O}_6\ (\text{fructose})

Water is in large excess and the acid is a catalyst, so the reaction behaves as first order in sucrose. A solution of $0.200\ \text{g mL}^{-1}$ sucrose in the acid is placed in a polarimeter tube $2.00\ \text{dm}$ long. The figure shows two readings of the angle of rotation: $+26.6^\circ$ at the start and $+16.0^\circ$ after $30.0\ \text{min}$ . A sugar of specific rotation $[\alpha]$ at concentration $c$ (in $\text{g mL}^{-1}$ ) turns the light by $\alpha = [\alpha]\,l\,c$ in a tube of length $l$ (in dm), and the angles of the sugars in a mixture add. Take $[\alpha] = +66.5^\circ$ for sucrose, $+52.5^\circ$ for glucose and $-92.4^\circ$ for fructose, and molar masses of $342.3\ \text{g mol}^{-1}$ for sucrose and $180.2\ \text{g mol}^{-1}$ for each product. Ignore any change of volume. Glucose and fructose do not react further in the acid during the run.

A graph of the angle of rotation shown by the polarimeter, in degrees, against time, in minutes, for the first 30 minutes of the run. Two readings are marked as points: plus 26.6 degrees at the start and plus 16.0 degrees at 30.0 minutes, joined by a falling curve.

What angle will the polarimeter show once all the sucrose has reacted?
What is the rate constant?
What is the half-life of sucrose in this acid?
After how long does the angle of rotation pass through zero?
What angle does the polarimeter show after $90.0\ \text{min}$ ?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Find the final angle from the products. Then the distance of each reading from that final angle falls exactly like the sucrose does.

(a) The end of the reaction. Sucrose: $\dfrac{0.200}{342.3} = 5.843 \times 10^{-4}\ \text{mol mL}^{-1}$ . So each product is $5.843 \times 10^{-4} \times 180.2 = 0.1053\ \text{g mL}^{-1}$ . The angles add:

\displaystyle \alpha_\infty = 2.00 \times 0.1053 \times (52.5 + (-92.4))^\circ = 2.00 \times 0.1053 \times (-39.9)^\circ = -8.40^\circ

The mixture turns the light the other way: the sugar has been inverted. (The start agrees with the figure: $66.5 \times 2.00 \times 0.200 = 26.6^\circ$ .)

(b) Rate constant. At any time, $\alpha_t - \alpha_\infty$ is proportional to the sucrose left. So

\displaystyle k = \frac{1}{30.0}\ln\frac{26.6 - (-8.40)}{16.0 - (-8.40)} = \frac{1}{30.0}\ln\frac{35.00}{24.40} = \frac{0.3607}{30.0}\ \text{min}^{-1} = 0.01202\ \text{min}^{-1}

(c) Half-life. $t_{1/2} = \dfrac{0.693}{0.01202}\ \text{min} = 57.6\ \text{min}$ .

(d) Zero rotation. $\alpha_t = 0$ means $\alpha_t - \alpha_\infty = 8.40^\circ$ , so $t = \dfrac{1}{k}\ln\dfrac{35.00}{8.40} = \dfrac{1.4269}{0.01202}\ \text{min} = 118.7\ \text{min}$ .

(e) After $90.0\ \text{min}$ . $\alpha = -8.40 + 35.00 \times e^{-1.0822} = -8.40 + 35.00 \times 0.3388 = 3.46^\circ$ , still a little to the right.

Check. Working out the mass of each sugar at each time and adding their three angles gives the same readings, and the same time for zero rotation.

Watch out. The angle itself does not fall like the sucrose, because the products turn the light too. Using $\ln\dfrac{\alpha_0}{\alpha_t}$ gives $k = 0.01694\ \text{min}^{-1}$ , which is wrong.

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