Consider the reaction.

2 upper C (s, graphite) plus upper H subscript 2 right arrow upper C subscript 2 upper H subscript 2 (g).

Standard Enthalpies of Formation
Substance
DHf (kJ/mol)
C2 H2 (g)
226.73
CaCO3 (s)
–1206.92
CaO (s)
–635.09
CO (g)
–110.525
CO2 (g)
–393.509
H 2 O (I)
–285.8
H 2 O (g)
–241.818
C (s), diamond
1.895
C (s), graphite
0.0

Based on the equation and the information in the table, what is the enthalpy of the reaction?
Use Delta H r x n equals the sum of delta H f of all the products minus the sum of delta H f of all the reactants..

–453.46 kJ
–226.73 kJ
226.73 kJ
453.46 kJ

226.73 kJ

Explanation:

Reaction:

2 C (s, graphite) + H2 (g) -> C2H2 (g)

The enthalpy of formation (ΔHf) of C2H2 (g) is 226.73 kJ/mol.

The enthalpy of formation of C (s, graphite) is zero because Carbon is in its elemental state.

The enthalpy of formation of H2 (g) is zero because is taken as a reference.

ΔHrx = ∑ΔHf of products - ∑ΔHf of reactants = 226.73 kJ/mol * 1 mol - 0 = 226.73 kJ

Note: 1 mol of C2H2 is formed.

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.