Answer
-1430 kJ
Explanation
Given information:
From 1; ΔH for the formation of 1 mol CO₂(g) = -394 kJ. For 2 mol CO₂(g), ΔH will be (2 x -394 kJ) = -788 kJ
From 2; ΔH for 1 mol H₂O(l) = -242 kJ. For 3 mol H₂O(l), ΔH will be (3 x -242 kJ) = -726 kJ
From 3, ΔH for 1 mol C₂H₆ = -84 kJ.
ΔHc for the given reaction can be calculated using the formula below:
[tex]ΔHc=\sum_^ΔH_f(products)-\sum_^ΔH_f(reactants)[/tex][tex]\begin{gathered} \sum_^ΔH_f(Products)=ΔH_f(2CO_2)+ΔH_f(3H_2O) \\ \\ \Rightarrow\sum_^ΔH_f(Products)=-788kJ+(-726kJ)= \\ \\ \sum_^ΔH_f(Products)=-788kJ-726kJ \\ \\ \sum_^ΔH_f(Products)=-1514kJ \end{gathered}[/tex]
For the reactants,
[tex]\begin{gathered} \sum_^ΔH_f(Reactants)=ΔH_f(C_2H_6)+ΔH_f(\frac{7}{2}O_2) \\ \\ \sum_^ΔH_f(Reactants)=+84+0=+84kJ \end{gathered}[/tex]
Therefore, ΔHc for the given reaction is:
ΔHc = -1514 kJ - (-84 kJ)
ΔHc = -1514kJ + 84 kJ
ΔHc = -1430 kJ
