Question

"Water gas" is an industrial fuel composed of a mixture of carbon monoxide and hydrogen gases. When this fuel is burned, carbon dioxide and water result. From the information given below, write a balanced equation and determine the enthalpy of this reaction:
CO(g) + ½ O2(g) → CO2(g) enthaplhy change + 282.8 kJ

H2(g) + ½ O2(g) → H2O(g) enthalphy change + 241.8 kJ

Answer 1

CO(g) + H₂(g) + O₂(g) → CO₂(g) + H₂O(g)

H_reaction = H₁ + H₂
H_reaction = 282.8 kJ + 241.8 kJ = 524.6 kJ

Note:
Enthaplhy change is usually negative value when it comes to exothermic processes. Enthaplhy change’s unit usually are kJ/mol. That’s why it is better to use this formulation of task: "Watergas" is an industrial fuel composed of a mixture of carbon monoxide and hydrogen gases. When this fuel is burned, carbon dioxide and water result. From the information given below, write abalanced equation and determine the enthalpy of this reaction:

CO(g) + ½O₂(g) → CO₂(g) enthaplhychange H₁= -282.8 kJ/mol
H₂(g) + ½O₂(g) → H₂O(g) enthalphychange H₂= -241.8 kJ/mol

And the solution is:

CO(g) + H₂(g) + O₂(g) → CO₂(g) + H₂O(g)

H_reaction = H₁ + H₂
H_reaction = -282.8 kJ/mol + (-241.8 kJ/mol) = -524.6 kJ/mol