Question

pls give me the answer immediately

Answer 1

GIVEN :

$\sf \bullet\:$ C(graphite) + O₂(g) ➝ CO₂(g) ∆ᵣH° = -393.51 kJ/mol

$\sf \bullet\:$ H₂(g) + (1/2)O₂(g) ➝ H₂O(l) ∆ᵣH° = -285.8 kJ/mol

$\sf \bullet\:$ CO₂(g) + 2H₂O(l) ➝ CH₄(g) + 2O₂(g) ∆ᵣH° = +890.3 kJ/mol

$\sf \bold{To\:find\::}$

C(graphite) + 2H₂(g) ➝ CH₄(g) ∆ᵣH° = ?

$\sf \bold{Solution\::}$

C(graphite) + O₂(g) ➝ CO₂(g) ∆ᵣH° = -393.51 kJ/mol .........1

H₂(g) + (1/2)O₂(g) ➝ H₂O(l) ∆ᵣH° = -285.8 kJ/mol. .........2

CO₂(g) + 2H₂O(l) ➝ CH₄(g) + 2O₂(g) ∆ᵣH° = +890.3 kJ/mol. ..........3

Multiply equation 2 by 2,

2× [H₂(g) + (1/2)O₂(g) ➝ H₂O(l)] ∆ᵣH° = 2 ×(-285.8 kJ/mol)

2H₂(g) + O₂(g) ➝ 2H₂O(l) ∆ᵣH° = -517.6 kJ/mol. ......... equation 4

Now add equation 1,2 and 4

C(graphite) + O₂(g) ➝ CO₂(g) ∆ᵣH° = -393.51 kJ/mol

CO₂(g) + 2H₂O(l) ➝ CH₄(g) + 2O₂(g) ∆ᵣH° = +890.3 kJ/mol

2H₂(g) + O₂(g) ➝ 2H₂O(l) ∆ᵣH° = -517.6 kJ/mol.

_______________________________________________

C(graphite) + 2O₂(g) + CO₂(g) + 2H₂O(l) + 2H₂(g) ➝ CO₂(g) + CH₄(g) + 2O₂(g) + 2H₂O(l) ∆ᵣH° = (-393.51 kJ/mol + 890.3 kJ/mol - 517.6 kJ/mol)

➝ C(graphite) + 2H₂(g) ➝ CH₄(g) ∆ᵣH° = -20.81 kJ/mol

$\sf \bold{ANSWER\::\:\:\:-20.81\:kJ/mol}$

Answer 2

Answer:

−20.81kJ/mol is the correct answer.