Question 6.6: Calculating ΔH from Calorimetric Data Suppose 0.562 g of gra...
Calculating ΔH from Calorimetric Data
Suppose 0.562 g of graphite is placed in a calorimeter with an excess of oxygen at 25.00°C and 1 atm pressure (Figure 6.13). Excess O_2 ensures that all carbon burns to form CO_2. The graphite is ignited, and it burns according to the equation
C(graphite) + O_2(g) → CO_2(g)On reaction, the calorimeter temperature rises from 25.00°C to 25.89°C. The heat capacity of the calorimeter and its contents was determined in a separate experiment to be 20.7 kJ/°C. What is the heat of reaction at 25.00°C and 1 atm pressure? Express the answer as a thermochemical equation.
PROBLEM STRATEGY
STEP 1 You obtain the heat released by the graphite sample when it burns in the calorimeter. You reason as follows. The heat released by the reaction is absorbed by the calorimeter and its contents. Let q_{rxn} be the quantity of heat from the reaction mixture, and let C_{cal} be the heat capacity of the calorimeter and contents. The quantity of heat absorbed by the calorimeter is C_{cal}Δt. This will have the same magnitude as q_{rxn}, but the opposite sign: q_{rxn} = –C_{cal}Δt. Substituting values of C_{cal} and Δt, you obtain the heat q_{rxn} from the sample of C (0.562 g). The factor (q_{rxn}/0.562 g C) converts
Grams C → kJ heat
STEP 2 The ΔH of reaction is the heat obtained from 1 mol C. Imagine you start with this amount of C and convert (mathematically) to grams C, then to kJ heat (ΔH for the reaction). The conversion steps are:
Moles C → grams C → kJ heat (ΔH)
STEP 1 The heat from the graphite sample is
\begin{aligned}q_{r x n} &= -C_{c a l} \Delta t = -20.7 kJ /{ }^{\circ} C \times \left(25.89^{\circ} C – 25.00^{\circ} C \right) \\ &=-20.7 kJ /{ }^{\circ} C \times 0.89^{\circ} C = -18.4 kJ\end{aligned}An extra figure is retained in q_{rxn} for further computation. The negative sign indicates the reaction is exothermic, as expected for a combustion. The factor to convert grams C to kJ heat is -18.4 kJ/0.562 g C.
STEP 2 The conversion of 1 mol C to kJ heat for 1 mol (ΔH) is
1 \cancel{mol C} \times \frac{12 \cancel{g C}}{1 \cancel{mol C}} \times \frac{-18.4 kJ }{0.562 \cancel{g C}} = -3.9 \times 10^2 kJ(The final answer has been rounded to two significant figures.) When 1 mol of carbon burns, 3.9 × 10² kJ of heat is released. You can summarize the results by the thermochemical equation:
C\text {(graphite)} + O _2(g) \longrightarrow CO _2(g) ; \Delta H = -3.9 \times 10^2 kJ