Calculate the standard enthalpy change for the reaction 2C₈H₁₈(l) + 21O₂(g) $\rarr$ 8CO(g) + 8CO₂(g) + 18H₂O(l) .Given:
2C₈H₁₈(l) + 25O₂(g) $\rarr$ 16CO₂(g) + 18H₂O(l) $\Delta$ H^$\circ$ = -11,020 kJ/mol
2CO(g) + O₂(g) $\rarr$ 2CO₂(g) $\Delta$ H^$\circ$ = -566.0 kJ/mol

Question

Calculate the standard enthalpy change for the reaction 2C₈H₁₈(l) + 21O₂(g)

\rarr

8CO(g) + 8CO₂(g) + 18H₂O(l) .Given:
2C₈H₁₈(l) + 25O₂(g)

\rarr

16CO₂(g) + 18H₂O(l)

\Delta

H^$\circ$ = -11,020 kJ/mol
2CO(g) + O₂(g)

\rarr

2CO₂(g)

\Delta

H^$\circ$ = -566.0 kJ/mol

Quizplus · Accepted Answer

The correct answer is found by using Hess's Law, which allows us to add or subtract given reactions to find the enthalpy change of the target reaction. The first given reaction has an enthalpy change of -11,020 kJ/mol. The second reaction, which forms 2 COS from 2 CO and O2, has an enthalpy change of -566.0 kJ/mol but needs to be multiplied by 8 (since the target reaction produces 8 COS) resulting in -4528 kJ/mol. Subtracting this from the first reaction's enthalpy change gives -11,020 kJ/mol - (-4528 kJ/mol) = -6492 kJ/mol, which matches option C. However, due to a mistake in the selection process, the correct mathematical operation and result align with the explanation but not with the initially indicated choice. The correct calculation aligns with choice C, not B, based on the provided explanation.

Calculate the Standard Enthalpy Change for the Reaction 2C8H18(l) →\rarr→ 8CO(g) + 8CO2(g) + 18H2O(l)

Calculate the Standard Enthalpy Change for the Reaction 2C₈H₁₈(l) $\rarr$ 8CO(g) + 8CO₂(g) + 18H₂O(l)