Calculate the standard enthalpy change for the reaction 2C₈H₁₈(l) + 21O₂(g) → 8CO(g) + 8CO₂(g) + 18H₂O(l) . Given:
2C₈H₁₈(l) + 25O₂(g) → 16CO₂(g) + 18H₂O(l) ΔH° = -11,020. kJ/mol
2CO(g) + O₂(g) → 2CO₂(g) ΔH° = -566.0 kJ/mol

Question

Quizplus · Accepted Answer

To find the standard enthalpy change for the given reaction, we can use the given reactions and their enthalpies to construct the desired reaction. The first given reaction is directly used in the calculation. The second reaction needs to be adjusted to fit the desired reaction's coefficients. Since the desired reaction produces 8 CO(g), we need to reverse the second given reaction and multiply it by 4 (to get 8 CO on the reactant side), which changes the sign of ΔH° and multiplies it by 4. Therefore, the total ΔH° for the desired reaction is the sum of -11,020 kJ/mol (from the first reaction) and 4 times 566 kJ/mol (since we reversed and multiplied the second reaction), resulting in -11,020 kJ/mol + (4 * 566 kJ/mol) = -11,020 kJ/mol + 2,264 kJ/mol = -8,756 kJ/mol.

Calculate the Standard Enthalpy Change for the Reaction 2C8H18(l)

Calculate the Standard Enthalpy Change for the Reaction 2C₈H₁₈(l)