Question 7.2: Heat of Reaction: Moles Methane undergoes combustion with O2......
Heat of Reaction: Moles
Methane undergoes combustion with O_2 according to the following equation:
CH_4(g) + 2 O_2(g) → CO_2(g) + 2 H_2O(l) ΔH = -213 \frac{kcal}{mol\ CH_4}
How much heat (in kcal and kJ) is released during the combustion of 0.35 mol of methane?
ANALYSIS Since the value of ΔH for the reaction (213 kcal/mol) is negative, it indicates the amount of heat released when 1 mol of methane reacts with O_2. We need to find the amount of heat released when an amount other than 1 mol reacts, using appropriate factor-label calculations to convert from our known or given units to kilocalories, and then to kilojoules.
BALLPARK ESTIMATE Since 213 kcal is released for each mole of methane that reacts, 0.35 mol of methane should release about one-third of 213 kcal, or about 70 kcal. There are about 4 kJ per kcal, so 70 kcal is about 280 kJ.
To find the amount of heat released (in kilocalories) by combustion of 0.35 mol of methane, we use a conversion factor of kcal/mol, and then we can convert to kilojoules using a kJ/kcal conversion factor (see Section 1.13):
0.35 \cancel{ mol\ CH_4} × \frac{-213\ kcal}{1\ \cancel{mol\ CH_4}} = -75 kcal
-75 \cancel{kcal} × \left(\frac{4.184\ kJ}{\cancel{kcal}} \right) = -314 kJ
The negative sign indicates that the 75 kcal (314 kJ) of heat is released.
BALLPARK CHECK The calculated answer is consistent with our estimate (70 kcal or 280 kJ).