Question 8 of 20 Using the equations H2(g) + F2 (g) → 2 HF (g) AH°...

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Answer 1

Answer #1

ANSWER:

Given, the equations are:

(1) H2 (g) + F2 (g) $\rightarrow$ 2 HF (g) $\Delta$ Ho = - 79.2 kJ/mol

(2) C (s) + 2 F2 (g) $\rightarrow$ CF4 (g) $\Delta$ Ho = 141.3 kJ/mol

(3) 2 C (s) + 2 H2 (g) $\rightarrow$ C2H4 (g) $\Delta$ Ho = - 97.6kJ/mol

On doing operation 2 x (1) + 2 x (2) - (3), we get

C2H4 (g) + 6 F2 (g) $\rightarrow$ 2 CF4 (g) + 4 HF (g)

So, enthalpy of reaction can be obtain by doing operation : 2 x (1) + 2 x (2) - (3). Now,

$\Delta$ Horxn = 2 x (- 79.2 kJ/mol) + 2 x (141.3 kJ/mol) - (- 97.6kJ/mol)

$\Delta$ Horxn = 221.8 kJ/mol

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Answer 2

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