Final answer:
To calculate ∆H for the formation of C₂H₂(g) from 2C(s) and H₂(g), we use Hess's Law, reversing and combining given reactions to match the target reaction, resulting in a ∆H of +226.8 kJ.
Step-by-step explanation:
To calculate the enthalpy change (∆H) for the reaction 2C(s) + H₂(g) → C₂H₂(g), we must manipulate given chemical equations to match the target reaction. We will use Hess's Law which states that the change in enthalpy for a reaction is the same no matter how many steps the reaction is carried out in. This means that if a series of reactions are added together, the overall ∆H is the sum of the ∆H values for each individual reaction.
First, we reverse the combustion of acetylene reaction to target the formation of acetylene:
C₂H₂(g) + O₂(g) ← 2CO₂(g) + H₂O(l); ∆H = +1299.6 kJ
Next, we consider the following reactions:
2C(s) + 2O₂(g) → 2CO₂(g); ∆H = -787.0 kJ(since 2C(s) + O₂(g) → CO₂(g); ∆H = -393.5 kJ is doubled)H₂(g) + ½O₂(g) → H₂O(l); ∆H = -285.8 kJ
By adding the above reactions, we can cancel out the O₂(g) and H₂O(l), leaving us with the target reaction:
2C(s) + H₂(g) → C₂H₂(g)
To obtain ∆H for this reaction, we add the ∆H values of the manipulated reactions:
1299.6 kJ (for the reversed C₂H₂ combustion) - 787.0 kJ (for the carbon combustion twice) - 285.8 kJ (for hydrogen combustion)
Doing the math gives us:
∆H = 1299.6 kJ - 787.0 kJ - 285.8 kJ = +226.8 kJ
The enthalpy change per mole for the formation of acetylene from carbon and hydrogen gas is +226.8 kJ.