Question

Calculate the standard enthalpy of formation of gaseous hydrogen fluoride (HF) using the following thermochemical information:

C2H4(g) + 6 F2(g) 2 CF4(g) + 4 HF(g) (delta)H = -2486.3 kJ

CF4(g) C(s) + 2 F2(g) (delta)H = +680 kJ

2 C(s) + 2 H2(g) C2H4(g) (delta)H = +52.3 kJ

(delta)H =__________ kJ

Answer 1

Dear student,

Answer of C₂ Hy(9) + 6 F2 (g) 2CF429) + 4HF (9) AH = -2 4 86.3KT CF4 (9) c(S) + 2F219) o AH = +680 kg 2 C(s) + 2 H2O) - C₂ Hy (9) - ③ OH = + 52.3kJ A Hp (HF(9)) 2 HF (9) H2(g) + F219) = -

Now , + ③ 2x ① = 0 + 2x 2x BHp CHF) = -2486.3 KJ + 2x680 KI. et 520.3 kJ 2xo Hp CHF) = -2486.34 1360+52.3 kJ 2x A Hf (HF) = -1074 KI o Hq (HF) = -1074 kJ A Hg (IFF) = ~587 kJ

Hence standard enthalpy of formation of HF (g) is -537 kj.

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