Answer to Question #140812 in General Chemistry for Bushra Aqila Binti Ramli

Question

Given the following enthalpies of the reactions, calculate the enthalpy change for the
oxidation of ethyne (C2H2).
C2H2(g) + 5/2 O2(g) → 2CO2(g)+ H2O(l)
C(s) + O2(g) → CO2(g)
H = -394 kJ/mol
2C(s) + H2(g) → C2H2(g)
H = +227 kJ/mol
H2(g) + ½ O2(g) → H2O(l)
H = -286 kJ/mol

Answer 1

Given the following enthalpies of the reactions, calculate the enthalpy change for the oxidation of ethylene (C2H2).

C₂H₂ (g) + 5/2 O₂ (g) ---> 2CO₂ (g) + H₂O(l) ---- Reaction 1

C(s) + O₂ (g) ----> CO₂ (g); ΔH = -394 kJ/mol ------ Reaction 2

2C(s) + H₂ (g) ------> C₂H₂ (g) ; ΔH = +227 kJ/mol ----- Reaction 3

H₂ (g) + 1/2 O₂ (g) -----> H₂O(l); ΔH = -286 kJ/mol ------ Reaction 4

Solution:

We are given enthalpies of the three reactions and asked to find enthalpy change for the oxidation of ethylene (C₂H₂).

Hess’s Law: Enthalpy change of a reaction is independent of the pathway of that reaction.

If a reaction can take place by more than one route then the enthalpy of the reaction is independent of the path and depends only on the initial and final condition of the reaction.

We can arrange the given three reactions in such a way that we get the reaction of oxidation of ethylene.

Let us look at the reactions

In reaction 1, CO₂ has coefficient 2, but in reaction 2 the coefficient of CO₂ is 1, so we multiply reaction 2 by 2.

In reaction 1, C₂H₂ is on the left side but in rection 3 it is on the right side, so we will

have to reverse the reaction 3.

In both reaction 1 and reaction 4, H₂O is on the right side, so there is no need to do any changes.

After doing all these operations we have

2C(s) + 2O₂ (g) ----> 2 CO₂ (g); ΔH = - 2* 394 kJ/mol ------ Reaction 2

C₂H₂ (g) ----> 2C(s) + H₂ (g); ΔH = - 227 kJ/mol ----- Reaction 3

H₂ (g) + 1/2 O₂ -----> (g) H₂O(l); ΔH = -286 kJ/mol ------ Reaction 4

Note that ΔH of reaction 2 is multiplied by 2, the sign of ΔH of reaction 3 is reversed and ΔH of reaction 4 is unchanged.

Now add reaction 2, 3 and 4

2C(s) + 2O₂ (g) ----> 2 CO₂ (g); ΔH = - 788 kJ/mol

C₂H₂ (g) ------> 2C(s) + H₂ (g); ΔH = - 227 kJ/mol

H₂ (g) + 1/2 O₂ (g) -----> H₂O(l); ΔH = -286 kJ/mol

C₂H₂ (g) + 2O₂ (g) + 1/2 O₂ (g) -------> 2 CO₂ (g) + H₂O(l); ΔH = -788 -227 – 286 = -1301 kJ/mol

Hence, C₂H₂ (g) + 5/2 O₂ (g) ------> 2 CO₂ (g) + H₂O(l); ΔH = -1301 kJ/mol