Answer to Question #204801 in Chemistry for Ankush

Question #204801

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)




1
Expert's answer
2021-06-09T12:16:44-0400

(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 ∆Hf for CH4 will be: −74.8kJ

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