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Multiple Choice
Given the heat of formation of TiI3(s) is -328 kJ/mol and the heat of reaction for 2Ti(s) + 3I2(g) → 2TiI3(s) is -839 kJ, what is the heat of sublimation of I2(s) at 25°C?
A
-383 kJ/mol
B
-183 kJ/mol
C
-83 kJ/mol
D
-283 kJ/mol
Verified step by step guidance
1
Understand the given values: The heat of formation of TiI3(s) is -328 kJ/mol, and the heat of reaction for 2Ti(s) + 3I2(g) → 2TiI3(s) is -839 kJ.
Recall that the heat of formation is the energy change when one mole of a compound is formed from its elements in their standard states. Here, TiI3(s) is formed from Ti(s) and I2(g).
Use Hess's Law, which states that the total enthalpy change for a reaction is the same, no matter how many steps the reaction is carried out in. This allows us to relate the heat of formation and the heat of reaction.
Set up the equation using Hess's Law: The heat of reaction for 2Ti(s) + 3I2(g) → 2TiI3(s) is given as -839 kJ. The heat of formation for 2 moles of TiI3(s) is 2 * (-328 kJ/mol) = -656 kJ.
Calculate the heat of sublimation of I2(s) by rearranging the equation: Heat of reaction = Heat of formation of products - Heat of formation of reactants. Solve for the heat of sublimation of I2(s) using the equation: -839 kJ = -656 kJ + 3 * (Heat of sublimation of I2(s)).