Problem: Use the ΔH°f and ΔH°rxn information provided to calculate ΔH°f for IF.IF7(g) + I2(g) → IF5(g) + 2IF(g)                        ΔH°rxn = -89kJ                ΔH°f(kJ/mol)IF7(g)              -941IF5(g)              -840a. 24kJ/molb. 101 kJ/molc. -95 kJ/mold. -190 kJ/mole. -146 kJ/mol

FREE Expert Solution

We're asked to calculate the ΔH°f for IF for the reaction: 


IF7(g) + I2(g)  IF5(g) + 2 IF(g) ΔH°rxn = -89 kJ


Recall that we can calculate the enthalpy of formation (ΔH˚f) of IF from the given ΔH˚rxn using the equation:


H°rxn=H°f,products-H°f, reactants


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Problem Details

Use the ΔH°f and ΔH°rxn information provided to calculate ΔH°f for IF.

IF7(g) + I2(g) → IF5(g) + 2IF(g)                        ΔH°rxn = -89kJ
                ΔH°f(kJ/mol)
IF7(g)              -941
IF5(g)              -840

a. 24kJ/mol

b. 101 kJ/mol

c. -95 kJ/mol

d. -190 kJ/mol

e. -146 kJ/mol

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