What is the standard enthalpy change (ΔH°) for the reaction: CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)?

• A. -393 kJ
• B. -571 kJ
• C. -890 kJ
• D. +890 kJ

Answer: (C)

Verified step by step guidance

1. Identify the reaction given: \( \mathrm{CH_4(g)} + 2 \mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)} + 2 \mathrm{H_2O(l)} \). We want to find the standard enthalpy change \( \Delta H^\circ \) for this reaction.
2. Recall that the standard enthalpy change of a reaction can be calculated using standard enthalpies of formation (\( \Delta H_f^\circ \)) of the reactants and products with the formula: \[ \Delta H^\circ = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants}) \]
3. Look up or use the given standard enthalpies of formation for each species involved: - \( \Delta H_f^\circ (\mathrm{CH_4(g)}) \) - \( \Delta H_f^\circ (\mathrm{O_2(g)}) = 0 \) (since it is an element in its standard state) - \( \Delta H_f^\circ (\mathrm{CO_2(g)}) \) - \( \Delta H_f^\circ (\mathrm{H_2O(l)}) \)
4. Multiply each \( \Delta H_f^\circ \) by its stoichiometric coefficient in the balanced equation, then sum the values for products and reactants separately.
5. Subtract the sum of the reactants' enthalpies from the sum of the products' enthalpies to find \( \Delta H^\circ \) for the reaction: \[ \Delta H^\circ = [\Delta H_f^\circ (\mathrm{CO_2(g)}) + 2 \times \Delta H_f^\circ (\mathrm{H_2O(l)})] - [\Delta H_f^\circ (\mathrm{CH_4(g)}) + 2 \times \Delta H_f^\circ (\mathrm{O_2(g)})] \]