A gaseous fuel mixture consisting of $3.67 ext{ L}$ of ethylene ($ ext{C}_2 ext{H}_4$) and methan

Question

A gaseous fuel mixture consisting of $3.67	ext{ L}$ of ethylene ($	ext{C}_2	ext{H}_4$) and methane ($	ext{CH}_4$) is subjected to complete combustion at $25^\circ	ext{C}$ and $1	ext{ atm}$ pressure, producing exactly $6.11	ext{ L}$ of carbon dioxide ($	ext{CO}_2$) gas measured under identical conditions. Calculate the amount of heat evolved (in $	ext{kJ}$) when exactly $1.00	ext{ L}$ of this initial gas mixture is burned under the same standard conditions. The standard enthalpies of combustion at $25^\circ	ext{C}$ are: $\Delta H^\circ_c[	ext{C}_2	ext{H}_4(g)] = -1423	ext{ kJ mol}^{-1}$ and $\Delta H^\circ_c[	ext{CH}_4(g)] = -891	ext{ kJ mol}^{-1}$. (Take the standard molar volume of an ideal gas at $25^\circ	ext{C}$ and $1	ext{ atm}$ to be $24.45	ext{ L mol}^{-1}$. Report your answer to one decimal place)

Professor PC Thomas Classes · Answer

Let $V_1$ be the volume of ethylene ($	ext{C}_2	ext{H}_4$) and $V_2$ be the volume of methane ($	ext{CH}_4$) in the initial mixture. We are given: $V_1 + V_2 = 3.67	ext{ L}$. The combustion equations are: $	ext{C}_2	ext{H}_4(g) + 3	ext{O}_2(g) ightarrow 2	ext{CO}_2(g) + 2	ext{H}_2	ext{O}(l)$ (1 volume of $	ext{C}_2	ext{H}_4$ yields 2 volumes of $	ext{CO}_2$), and $	ext{CH}_4(g) + 2	ext{O}_2(g) ightarrow 	ext{CO}_2(g) + 2	ext{H}_2	ext{O}(l)$ (1 volume of $	ext{CH}_4$ yields 1 volume of $	ext{CO}_2$). The total volume of $	ext{CO}_2$ produced is: $2V_1 + V_2 = 6.11	ext{ L}$. Subtracting the two volume equations gives: $(2V_1 + V_2) - (V_1 + V_2) = 6.11 - 3.67 \implies V_1 = 2.44	ext{ L}$. Therefore, $V_2 = 3.67 - 2.44 = 1.23	ext{ L}$. In $1.00	ext{ L}$ of the mixture, the volume fraction of ethylene is $v_1 = 2.44 / 3.67 = 0.66485	ext{ L}$, and that of methane is $v_2 = 1.23 / 3.67 = 0.33515	ext{ L}$. At $25^\circ	ext{C}$ and $1	ext{ atm}$, the number of moles in $1.00	ext{ L}$ is $n_{	ext{total}} = 1.00	ext{ L} / 24.45	ext{ L mol}^{-1} = 0.04090	ext{ mol}$. The moles of each gas in $1	ext{ L}$ are: $n(	ext{C}_2	ext{H}_4) = 0.66485 / 24.45 = 0.02719	ext{ mol}$, and $n(	ext{CH}_4) = 0.33515 / 24.45 = 0.01371	ext{ mol}$. The total heat evolved per liter of mixture is: $q = n(	ext{C}_2	ext{H}_4) 	imes |\Delta H^\circ_c(	ext{C}_2	ext{H}_4)| + n(	ext{CH}_4) 	imes |\Delta H^\circ_c(	ext{CH}_4)| = 0.02719 	imes 1423 + 0.01371 	imes 891 = 38.69 + 12.22 = 50.91	ext{ kJ} \approx 50.9	ext{ kJ}$.