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Multiple Choice
Using the information below, calculate the enthalpy change (ΔH°rxn) for the following reaction: CH4 (g) + 4 Cl2 (g) → CCl4 (g) + 4 HCl (g). Given the standard enthalpy of formation (ΔHf°) values: CH4 (g) = -75 kJ/mol, CCl4 (g) = -96 kJ/mol, and HCl (g) = -92 kJ/mol.
A
-390 kJ/mol
B
-450 kJ/mol
C
-270 kJ/mol
D
-150 kJ/mol
Verified step by step guidance
1
Identify the formula for calculating the enthalpy change of a reaction using standard enthalpies of formation: ΔH°rxn = ΣΔHf°(products) - ΣΔHf°(reactants).
List the standard enthalpy of formation values given: CH4 (g) = -75 kJ/mol, CCl4 (g) = -96 kJ/mol, HCl (g) = -92 kJ/mol.
Calculate the total enthalpy of formation for the products: CCl4 (g) and 4 HCl (g). Use the formula: ΣΔHf°(products) = ΔHf°(CCl4) + 4 * ΔHf°(HCl).
Calculate the total enthalpy of formation for the reactants: CH4 (g) and 4 Cl2 (g). Note that Cl2 is an element in its standard state, so its ΔHf° is 0. Use the formula: ΣΔHf°(reactants) = ΔHf°(CH4) + 4 * ΔHf°(Cl2).
Subtract the total enthalpy of formation of the reactants from the total enthalpy of formation of the products to find ΔH°rxn: ΔH°rxn = ΣΔHf°(products) - ΣΔHf°(reactants).