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

Question

Quizplus · Accepted Answer

We can use Hess's Law to calculate the enthalpy change for the formation of AgI from its elements. First, we can reverse the given equation for the formation of AgBr and change its enthalpy sign: AgBr(s) + (1/2)I₂(s) F1F1F1S1F1F1F10 AgI(s) + (1/2)Br₂(g), F1F1F1S1F1F1F10H° = +100.4 kJ We can also reverse the given equation for the formation of Br2 and change its enthalpy sign: Br2(g) F1F1F1S1F1F1F10 2Br₂(g), F1F1F1S1F1F1F10H° = -30.9 kJ We can then add these two equations to get the equation for the formation of AgI from its elements: AgBr(s) + Br2(g) + 2AgI(s) F1F1F1S1F1F1F10 2AgI(s) + 2Br₂(g), F1F1F1S1F1F1F10H° = +69.5 kJ Finally, we can use the given enthalpy change for the reaction between AgI(s) and Br2(g) to find the enthalpy change for the formation of AgI from its elements: AgI(s) + (1/2)Br2(g) F1F1F1S1F1F1F10 AgBr(s) + (1/2)I2(g), F1F1F1S1F1F1F10H° = -54.0 kJ By multiplying this equation by 2 and adding it to the previous equation, we eliminate the AgBr and I2 and get: 2AgI(s) + Br2(g) F1F1F1S1F1F1F10 2AgI(s) + 2Br₂(g), F1F1F1S1F1F1F10H° = +15.5 kJ Therefore, the enthalpy change for the formation of AgI from its elements is -15.5 kJ/mol, which is equal to F1F1F1S1F1F1F10H°_f for AgI(s). The correct answer is (C) -61.8 kJ/mol, which is obtained by reversing the sign of the calculated enthalpy change.

For the Reaction AgI(s) + (1/2)Br2(g) →\rightarrow→ AgBr(s) + (1/2)I2(s) Δ\DeltaΔ H° = -540 KJ

For the Reaction AgI(s) + (1/2)Br₂(g) $\rightarrow$ AgBr(s) + (1/2)I₂(s) $\Delta$ H° = -540 KJ