Question

working with enthalpies of reaction and combustion
the enthalpy of combustion for octane (c₈h₁₈(l)), a key component of gasoline, is - 5074 kj/mol. the reaction equation is: c₈h₁₈(l)+12.5o₂(g)→8co₂(g)+9h₂o(g).
what is the δh for this reaction?
16co₂(g)+18h₂o(g)→2c₈h₁₈(l)+25o₂(g)
kj

Explanation:

Step1: Recall enthalpy property

The enthalpy change of a reverse - reaction has the opposite sign of the forward - reaction.

Step2: Analyze given reactions

The first reaction is $C_8H_{18}(l)+12.5O_2(g)
ightarrow8CO_2(g) + 9H_2O(g)$ with $\Delta H=- 5074\ kJ/mol$. The second reaction $16CO_2(g)+18H_2O(g)
ightarrow2C_8H_{18}(l)+25O_2(g)$ is the reverse of the first reaction multiplied by a factor of 2.

Step3: Calculate $\Delta H$

If the first reaction has $\Delta H=-5074\ kJ/mol$, then for the reverse reaction of the first one (multiplied by 2), $\Delta H = 2\times5074\ kJ$.

Answer:

$10148$