smc-4266-50-Calorimetry

CHEMISTRY, M4 EQ-Bank 25

In an experiment, 1.40 g of pure sucrose, \(\ce{C12H22O11}\), underwent complete combustion, heating 350 mL of water from 19.6 °C to 35.8 °C.

\(\ce{MM(C12H22O11)}\) = 342 g mol\(^{−1}\)

Assuming there was no heat lost to the surroundings, calculate the experimental heat of combustion of pure sucrose, in joules per gram. (2 marks)

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\(1.65 \times 10^{4}\ \text{J g}^{-1} \)

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\(\text{Energy absorbed} \ce{(H2O)}\ = mC \Delta T = 0.350 \times 4.18 \times 10^{3} \times (35.8-19.6) = 23\ 070.6\ \text{J} \)

\(\text{Energy (sucrose)}\ = \dfrac{23\ 070.6}{1.40} = 1.65 \times 10^{4}\ \text{J g}^{-1} \)

The heat of combustion of a sample of crude oil is to be determined using a bomb calorimeter. All of the students in a class are given the same method to follow. The apparatus used by the students is shown below.

For this experiment, the students could maximise

precision by using a digital thermometer \(\pm\)0.2 °C.
validity by calculating the heat of combustion per mole.
accuracy by taking samples from three different sources.
uncertainty by having all students closely follow the same experimental procedure.

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\(\A\)

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Consider each option:

→ Option A: Error involved in reading the thermometer would be improved by this measure.

→ Option B: Energy per mole is inappropriate as crude oil is used.

→ Option C: Accuracy dependent on process technique, not the source of samples.

→ Option D: If the procedure itself creates inaccuracy, this is incorrect.

\(\Rightarrow A\)

♦ Mean mark 49%.

CHEMISTRY, M4 EQ-Bank 30

A student used the apparatus shown to determine the molar heat of combustion of ethanol.

The following results were obtained.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex}\text{Initial mass of burner}\rule[-1ex]{0pt}{0pt} & \text{133.20 g} \\
\hline
\rule{0pt}{2.5ex}\text{Final mass of burner}\rule[-1ex]{0pt}{0pt} & \text{132.05 g}\\
\hline
\rule{0pt}{2.5ex}\text{Initial temperature of water}\rule[-1ex]{0pt}{0pt} & \text{25.0°C}\\
\hline
\rule{0pt}{2.5ex}\text{Final temperature of water}\rule[-1ex]{0pt}{0pt} & \text{45.5°C}\\
\hline
\end{array}

Calculate ethanol's molar heat of combustion from this data? (3 marks)

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\(1030\ \text{kJ mol}^{-1} \)

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\(\ce{Ethanol \ \Rightarrow \ C2H6O }\)

\(\ce{m(C2H6O) = 133.20-132.05=1.15\ \text{grams}}\)

\(\ce{MM(C2H6O) = 2 \times 12.01 + 6 \times 1.008 + 16 = 46.068\ \text{g mol}^{-1}} \)

\(\ce{n(C2H6O) = \dfrac{\text{m}}{\text{MM}} = \dfrac{1.15}{46.068} = 0.02496\ \text{mol}}\)

\(\Delta H = mC\Delta T = 0.300 \times 4.18 \times 10^{3} \times (45.5-25) = 25\ 707\ \text{J} = 25.707\ \text{kJ} \)

\(\ce{C2H6O \text{molar heat of combustion}}\)

\(=\dfrac{25.707}{0.02496}\)

\(=1030\ \text{kJ mol}^{-1} \)