Given the reaction: CH_4(g) + 4 Cl_2(g) → CCl_4(l) + 4 HCl(g), what is the standard enthalpy change (ΔH°) for this reaction? Use the following standard enthalpies of formation: ΔH_f°(CH_4(g)) = -74.8 kJ/mol, ΔH_f°(Cl_2(g)) = 0 kJ/mol, ΔH_f°(CCl_4(l)) = -139.3 kJ/mol, ΔH_f°(HCl(g)) = -92.3 kJ/mol.

• A. +437.7 kJ
• B. +214.1 kJ
• C. -437.7 kJ
• D. -214.1 kJ

Answer: (C)

Verified step by step guidance

1. Write down the balanced chemical equation: \(\mathrm{CH_4(g) + 4\ Cl_2(g) \rightarrow CCl_4(l) + 4\ HCl(g)}\).
2. Recall the formula for the standard enthalpy change of a reaction using standard enthalpies of formation: \(\Delta H^\circ_{\text{reaction}} = \sum \Delta H^\circ_f (\text{products}) - \sum \Delta H^\circ_f (\text{reactants})\).
3. List the given standard enthalpies of formation: \(\Delta H^\circ_f (\mathrm{CH_4(g)}) = -74.8\ \mathrm{kJ/mol}\), \(\Delta H^\circ_f (\mathrm{Cl_2(g)}) = 0\ \mathrm{kJ/mol}\), \(\Delta H^\circ_f (\mathrm{CCl_4(l)}) = -139.3\ \mathrm{kJ/mol}\), \(\Delta H^\circ_f (\mathrm{HCl(g)}) = -92.3\ \mathrm{kJ/mol}\).
4. Calculate the sum of the enthalpies of formation for the products: \(\Delta H^\circ_f (\mathrm{CCl_4(l)}) + 4 \times \Delta H^\circ_f (\mathrm{HCl(g)})\).
5. Calculate the sum of the enthalpies of formation for the reactants: \(\Delta H^\circ_f (\mathrm{CH_4(g)}) + 4 \times \Delta H^\circ_f (\mathrm{Cl_2(g)})\), then subtract this from the products sum to find \(\Delta H^\circ_{\text{reaction}}\).