Use the ΔH°f and ΔH°rxn information provided to calculate ΔH°f for IF: ΔH°f (kJ/mol) IF7(g) + I2(g) → IF5(g) + 2 IF(g) ΔH°rxn = -89 kJ IF7(g) -941 IF5(g) -840 I2(g) 62.42

ΔH°f{IF(g)} = -127.58 kJ/mol

Explanation:

Hello.

In this case, since the information about the chemical reaction is:

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

ΔH°f{IF7(g)} = -941 kJ / mol

ΔH°f{IF5(g)} = -840 kJ / mol

ΔH°f{I2(g)} =62.42 kJ / mol (it should be 0 since I2 is a pure element that is nor formed by other elements, however, we will use the given value).

Since we don't know the enthalpy of formation of IF, but the enthalpy of reaction for this one is:

ΔH°rxn = ΔH°f{IF5(g)} + ΔH°f{IF(g)} - ΔH°f{IF7(g)} - ΔH°f{I2(g)}

We can obtain it via:

ΔH°f{IF(g)} = ΔH°rxn - ΔH°f{IF5(g)} + ΔH°f{IF7(g)} + ΔH°f{I2(g)}

ΔH°f{IF(g)} = -89 - (-840) + (-941) + 62.42

ΔH°f{IF(g)} = -127.58 kJ/mol

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