Below are I think the data for this problem:
Given the following data:
Ca (s) + 2 C (graphite) → CaC2 (s) ∆H = -62.8 kJ
Ca (s) + ½ O¬2 (g) → CaO (s) ∆H = -635.5 kJ
CaO (s) + H2O (l) → Ca(OH)2 (aq) ∆H = -653.1 kJ
C2H2 (g) + 5/2 O¬2 (g) → 2 CO2 (g) + H2O (l) ∆H = -1300 kJ
C (graphite) + O¬2 (g) → CO2 (g) ∆H = -393.51 kJ
Below is the answer:
CaC2 (s) + 2 H2O (l) → Ca(OH)2 (aq) + C2H2 (g)
So what you do is:
Times the first equation by -1 Second by 1 Third By 1 Fourth by -1 and Fifth by 2
So This gives us:
1.CaC2--> Ca+2C
2.Ca+1/2O2-->CaO
3.CaO+H2O-->Ca(OH)2
4.2CO2+H2O-->C2H2+5/2O2
5.2C+202-->2CO2
Now you cancel out like terms on either sides of the equation and you end up with
CaC2 (s) + 2 H2O (l) → Ca(OH)2 (aq) + C2H2 (g) Just what you wanted
So to calculate ∆H:
62.8-635.5-653.1+1300-787.02= -712.82
