Question

When C_2H_2 is burned in air, carbon dioxide and water are formed in the following reaction: C_2H_2+O_2(g) right arrow CO_2(g) + H_2O(l) delta H =-1300. kJ C(s) + O_2(g) right arrow CO_2(g)delta H =-394. kJ; H2(g) +1/2 O_2(g)right arrow H_2O(l) delta H =-286. kJ; Calculate delta H for 2C(s) + H_2(g) right arrow C_2H_2(g) Delta H = what is the heat of formation of H_2o(l) deltaHf

Answer 1

Given equations are :

C₂H₂(g) + (5/2)O₂(g)   $\rightarrow$ 2CO₂(g) + 1H₂O (l)   : $\Delta$ H₁ = -1300 kJ        ----(1)

C(s) + O₂(g) $\rightarrow$ CO₂(g)   : $\Delta$ H₂ = -394 kJ      ----(2)

H₂(g) + (1/2) O₂ (g) $\rightarrow$ H₂O(l) : $\Delta$ H₃ = -286 kJ                               ----(3)

2C(s) + H₂(g) $\rightarrow$ C₂H₂(g) :   $\Delta$ H = ?                                             ----(4)

Eqn(4) can be obtained from first three equations as follows:

Eqn(4) = reverse of Eqn(1) + [ 2xEqn (2) ] + Eqn (3)

So $\Delta$ H = - $\Delta$ H₁ + [2x $\Delta$ H₂] + $\Delta$ H₃

             = - (-1300 kJ ) + [2x(-394 kJ )] + (-286 kJ)

            = +226 kJ

-------------------------------------------------------------------------------------------

Heat of formation of H₂O (l) is the formation of 1 mole of H₂O (l) from its constituents elements.

H₂(g) + (1/2) O₂ (g) $\rightarrow$ H₂O(l) : $\Delta$ H₃ = -286 kJ                     

Here 1 mole of H₂O(l) is formed from its constituent elements H₂(g) & O₂(g) so in this case enthalpy change is -286 kJ

Therefore the enthalpy of formation of H₂O(l) is $\Delta$ H_f = -286 kJ