Answer to Question #204801 in Chemistry for Ankush

Question

From the following enthalpy changes,

C(s)  + O2(g) 🡪 CO2(g) ΔH  =  -393.5 kJ

H2(g)    +  ½ O2(g) 🡪 H2O(l) ΔH = -285.8 kJ 

CH4(g)    +   2O2(g)  🡪 CO2(g)  +  2H2O(l) ΔH  =  -890.3 kJ

calculate the value of ∆Hf for CH4. (3 marks for showing how to get to the overall reaction, stating the overall reaction  since  it  isn’t  given, and the heat of reaction)

Answer 1

(i) C(s) + O₂(g) ⟶ CO₂(g) ΔH = −393.5kJ

(ii) H₂(g)  + ½O2(g) ⟶ H₂O(l) ΔH = -285.8 kJ 

(iii) CH₄(g)  + 2O₂(g) ⟶ CO₂(g) + 2H₂O(l) ΔH = -890.3 kJ

Multiply (ii) by 2

(iv) 2H₂(g)  +  O2(g) ⟶ 2H₂O(l) ΔH = -285.8 kJ * 2 = -571.6kJ

Reverse (iii)

(v) CO₂​(g) + 2H₂​O(l)​ ⟶ CH₄(g) + 2O₂(g) ΔH = +890.3kJ

Add (i), (iv), (v)

(vi) C(s) + 2H₂(g) ​⟶ CH₄(g) ΔH = −74.8kJ

Hence, value of ∆H_f for CH₄ will be: −74.8kJ