Answer:
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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