Calculate DH for the reaction
2C(s) + H2(g) C2H2(g)
given the following chemical equations and their respective enthalpy changes:
C2H2(g) + O2(g) 2CO2(g) + H2O DH = -1299.6kJ
C(s) + O2(g) CO2(g) DH = -393.5kJ
H2(g) + O2(g) H2O(l) DH = -285.8kJ
Please help!!! I am completely lost! How do I start, what are the steps?
2CO2(g) + H2O(l) ----> C2H2(g) + 5/2O2(g) DH = + 1299.6kJ
2C(s) + 2O2(g) ----> 2CO2(g) DH = - 787kJ
H2(g) + O2(g) ----> H2O(l) DH = - 285.8kJ
By adding the above three equations
2C(s) + H2(g) ----> C2H2(g) DH = + 226.8 KJ
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