Question

Calculate the standard enthalpy of formation of gaseous carbon tetrafluoride (CF₄) using the following thermochemical information:

2 HF(g) H2(g) + F2(g)		H = +537 kJ
2 CF4(g) + 4 HF(g) C2H4(g) + 6 F2(g)		H = +2486.3 kJ
C2H4(g) 2 C(s) + 2 H2(g)		H = -52.3 kJ

Calculate the standard enthalpy of formation of gaseous carbon tetrafluoride (CF₄) using the following thermochemical information:

2 HF(g) H2(g) + F2(g)		H = +537 kJ
2 CF4(g) + 4 HF(g) C2H4(g) + 6 F2(g)		H = +2486.3 kJ
C2H4(g) 2 C(s) + 2 H2(g)		H = -52.3 kJ

H = kJ

Answer 1

$\Delta$H = -680 kJ

Explanation

Reaction 1 : 2 HF (g) $\rightleftharpoons$ H₂ (g) + F₂ (g) : $\Delta$ H₁ = 537 kJ

Reaction 2 : 2 CF₄ (g) + 4 HF (g) $\rightleftharpoons$ C₂H₄ (g) + 6 F₂ (g) : $\Delta$ H₂ = 2486.3 kJ

Reaction 3 : C₂H₄ (g) $\rightleftharpoons$ 2 C (s) + 2 H₂ (g) : $\Delta$ H₃ = -52.3 kJ

Desired reaction : C (s) + 2 F₂ (g) $\rightleftharpoons$ CF₄ (g)

Consider the operation : [2 * (Reaction 1) - (Reaction 2) - (Reaction 3)] / 2

This operation gives the desired reaction

$\Delta$H = [2 * ($\Delta$H₁) - ($\Delta$H₂) - ($\Delta$H₃)] / 2

$\Delta$H = [2 * (537 kJ) - (2486.3 kJ) - (-52.3 kJ)]

$\Delta$H = (1074 kJ - 2486.3 kJ + 52.3 kJ) / 2

$\Delta$H = (-1360 kJ) / 2

$\Delta$H = -680 kJ