For the reaction AgI(s) + (1/2)Br₂(g) → AgBr(s) + (1/2)I₂(s), ΔH° = -54.0 kJ ΔH°_f for AgBr(s) = -100.4 kJ/mol ΔH°_f for Br₂(g) = +30.9 kJ/mol The value of ΔH°_f for AgI(s) is

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Quizplus · Accepted Answer

Using Hess's Law, we can add the enthalpies of formation of the products and subtract the enthalpies of formation of the reactants. ΔH° = ΣΔH°_f(products) - ΣΔH°_f(reactants) We are given the values for ΔH°_f for AgBr(s) and Br₂(g), but not for AgI(s). However, we know that AgI(s) is a reactant in the equation, so we can use algebra to solve for its enthalpy of formation: ΔH°_f for AgI(s) = ΣΔH°_f(reactants) - ΣΔH°_f(products) ΔH°_f for AgI(s) = [-54.0 kJ/mol] - [(1/2)(-100.4 kJ/mol)] - [(1/2)(30.9 kJ/mol)] ΔH°_f for AgI(s) = -61.8 kJ/mol Therefore, the best choice is C, with an explanation that we can use Hess's Law and the given enthalpies of formation to solve for ΔH°_f for AgI(s).

For the Reaction AgI(s) + (1/2)Br2(g) → AgBr(s) + (1/2)I2(s), ΔH° = -54.0

For the Reaction AgI(s) + (1/2)Br₂(g) → AgBr(s) + (1/2)I₂(s), ΔH° = -54.0