Exploring the bomb calorimeter

Chemistry Level 2

A $0.8650 \text{ g}$ sample of lactic acid $(HC_{3}H_{5}O_{3})$ was burned in a bomb calorimeter. The chemical equation below describes the reaction that took place in the calorimeter.

$HC_{3}H_{5}O_{3(s)} + 3O_{2(g)} \longrightarrow 3CO_{2(g)} + 3H_2O_{(l)} \quad \Delta H^{\circ}_{rxn} = -1338 \space kJ/mol$

What is the temperature increase $($ in $^{\circ}C)$ in the calorimeter if the heat capacity of the calorimeter, including water, was determined to be $4.182 \space kJ/^{\circ}C$ ? Enter your answer to $3$ decimal places.

Take the molar mass of hydrogen, carbon, and oxygen to be $1.008 \text{ g}$ , $12.01 \text{ g}$ , and $16.00 \text{ g}$ , respectively.

The answer is 3.072.

1 solution

Akeel Howell
Apr 14, 2017

The molar mass of $HC_{3}H_{5}O_{3(s)}$ is $(1.008 + 3(12.01) + 5(1.008) + 3(16.0)) \text{ g}/mol = 90.078 \text{ g}/mol$ $0.8650 \text{ g}$ was burned $\implies \dfrac{0.8650 \text{ g}}{90.078 \text{ g}/mol} = 0.0096028 \ mol \ HC_3H_5O_{3(s)}$ burned.

Since $\Delta H^{\circ} = -1338 \space kJ/mol \ \implies \Delta H^{\circ} = \dfrac{-1338 \ kJ}{mol \space HC_3H_5O_{3(s)}} \times 0.0096028 \space mol \space HC_3H_5O_{3(s)} \\ = -12.8485464 \ kJ$

Since this is the energy change in the lactic acid (a loss of energy), it is also the energy absorbed by the calorimeter in which it is contained.

$\therefore$ The increase in temperature in the calorimeter is $\dfrac{12.8485464 \ kJ}{4.182 \ kJ/^{\circ}C} = 3.0723449^{\circ}C \approx 3.072^{\circ}C$ ( $3$ decimal places).

Exploring the bomb calorimeter

The answer is 3.072.

1 solution

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