Question 8.6: USING HESS’S LAW TO CALCULATE ΔH° Methane, the main constitu...
USING HESS’S LAW TO CALCULATE ΔH°
Methane, the main constituent of natural gas, burns in oxygen to yield carbon dioxide and water:
CH_{4}(g) + 2 O_{2}(g) → CO_{2}(g) + 2 H_{2}O(l)Use the following information to calculate ΔH° in kilojoules for the combustion of methane:
CH_{4}(g) + O_{2}(g) → CH_{2}O(g) + H_{2}O(g) ΔH° = -275.6 kJ
CH_{2}O(g) + O_{2}(g) → CO_{2}(g) + H_{2}O(g) ΔH° = -526.7 kJ
H_{2}O(l) → H_{2}O(g) ΔH° = 44.0 kJ
STRATEGY
It often takes some trial and error, but the idea is to combine the individual reactions so that their sum is the desired reaction. The important points are that:
• All the reactants [CH_{4}(g) and O_{2}(g)] must appear on the left.
• All the products [CO_{2}(g) and H_{2}O(l)] must appear on the right.
• All intermediate products [CH_{2}O(g) and H_{2}O(g)] must occur on both the left and the right so that they cancel.
• A reaction written in the reverse of the direction given [H_{2}O(g) → H_{2}O(l)] must have the sign of its ΔH° reversed (Section 8.6).
• If a reaction is multiplied by a coefficient [H_{2}O(g) → H_{2}O(l) is multiplied by 2], then ΔH° for the reaction must be multiplied by that same coefficient.
CH_{4}(g) + O_{2}(g) → \cancel{CH_{2}O(g)} + \cancel{H_{2}O(g)} ΔH° = -275.6 kJ
\cancel{CH_{2}O(g)} + O_{2}(g) → CO_{2}(g) + \cancel{H_{2}O(g)} ΔH° = -526.7 kJ
\underline{\cancel{2 [H_{2}O(g)} → H_{2}O(l)] 2 [ΔH° = -44.0 kJ] = -88.0 kJ}CH_{4}(g) → 2 O_{2}(g) → CO_{2}(g) + 2 H_{2}O(l) ΔH° = -890.3 kJ