Question

The enthalpy change for the reaction of hydrogen gas with fluorine gas to produce hydrogen fluoride is -542 kJ for the equation as written: H2(g) + F2(g) → 2 HF(g) ΔH = -542 kJ (a) What is the enthalpy change per mole of hydrogen fluoride produced?

Answer 1

- 271kJ

Step-by-step explanation:

The equation is given as;

H2(g) + F2(g) → 2 HF(g) ΔH = -542 kJ

In the equation above, the enthalpy change (-542kJ) is for the 2 moles of HF produced.

The question asked for the enthalpy change per mole(One mole) . So what we need to do is divide the enthalpy value of 2 moles by 2.

Hence we have;

542kJ = 2 moles

x = 1 mole

x = (542* 1) / 2

x = - 271kJ

Answer 2

-271 kJ per mole of hydrogen fluoride produced

Step-by-step explanation:

-542 kJ per mole relates with the balanced chemical equation which is the one with the lowest stoichiometric coefficients, then if you see the chemical equation you see that 1 mole of H2 and 1 mole of F2 react, then the enthalpy change of -542 kJ per mole is referring to the reaction of 1 mole of H2 with 1 mole of F2 to form 2 moles of HF. If you need the enthalpy change for just 1 of it you consider the reaction of 1/2 mole of H2 with 1/2 mole of F2, therefore you dont have initial proportion and then you divide the original enthalpy change by 2.