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