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Multiple Choice
Given the reaction IF7(g) + I2(g) → IF5(g) + 2 IF(g) with ΔH°rxn = -89 kJ, ΔH°f for IF7(g) = -941 kJ/mol, and ΔH°f for IF5(g) = -840 kJ/mol, calculate the standard enthalpy of formation (ΔH°f) for IF(g).
A
-150 kJ/mol
B
-100 kJ/mol
C
-200 kJ/mol
D
-123 kJ/mol
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Verified step by step guidance
1
Start by understanding the given reaction: IF7(g) + I2(g) → IF5(g) + 2 IF(g). This is a decomposition reaction where IF7 and I2 are reactants, and IF5 and IF are products.
Use Hess's Law, which states that the change in enthalpy for a reaction is the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants.
Write the expression for the standard enthalpy change of the reaction (ΔH°rxn) using the enthalpies of formation: ΔH°rxn = [ΔH°f(IF5) + 2 * ΔH°f(IF)] - [ΔH°f(IF7) + ΔH°f(I2)].
Substitute the known values into the equation: ΔH°rxn = -89 kJ, ΔH°f(IF7) = -941 kJ/mol, and ΔH°f(IF5) = -840 kJ/mol. Note that ΔH°f(I2) is zero because it is in its standard state.
Solve for ΔH°f(IF) by rearranging the equation: ΔH°f(IF) = [(ΔH°rxn + ΔH°f(IF7) - ΔH°f(IF5)) / 2]. This will give you the standard enthalpy of formation for IF(g).