Question

Study the reactions for the formation of compounds from their elements

Answer 1

-1430 kJ

Step-by-step 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:

`ΔHc=\sum_^ΔH_f(products)-\sum_^ΔH_f(reactants)`
`\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}`

For the reactants,

`\begin{gathered} \sum_^ΔH_f(Reactants)=ΔH_f(C_2H_6)+ΔH_f((7)/(2)O_2) \\ \\ \sum_^ΔH_f(Reactants)=+84+0=+84kJ \end{gathered}`

Therefore, ΔHc for the given reaction is:

ΔHc = -1514 kJ - (-84 kJ)

ΔHc = -1514kJ + 84 kJ

ΔHc = -1430 kJ