Given the thermochemical equation: CH4(g) + 2 H2(g) → C2H6(g), calculate ΔH° for the reaction if the standard enthalpies of formation are: ΔH°f(CH4(g)) = -74.8 kJ/mol, ΔH°f(H2(g)) = 0 kJ/mol, and ΔH°f(C2H6(g)) = -84.7 kJ/mol.

• A. +159.5 kJ
• B. -9.9 kJ
• C. -159.5 kJ
• D. +9.9 kJ

Answer: (B)

Verified step by step guidance

1. Identify the reaction and the given standard enthalpies of formation (ΔH°f) for each species involved: CH4(g), H2(g), and C2H6(g).
2. Recall that the standard enthalpy change for the reaction (ΔH°) can be calculated using the formula: \[\Delta H^\circ = \sum \Delta H^\circ_f (\text{products}) - \sum \Delta H^\circ_f (\text{reactants})\]
3. Write the expression for this specific reaction: \[\Delta H^\circ = \Delta H^\circ_f (\mathrm{C_2H_6(g)}) - \left[ \Delta H^\circ_f (\mathrm{CH_4(g)}) + 2 \times \Delta H^\circ_f (\mathrm{H_2(g)}) \right]\]
4. Substitute the given values into the equation: \[\Delta H^\circ = (-84.7\ \mathrm{kJ/mol}) - \left[ (-74.8\ \mathrm{kJ/mol}) + 2 \times 0 \right]\]
5. Simplify the expression to find the enthalpy change for the reaction, which will give you the value of ΔH°.