Question

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

Answer 1

Answer:

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.

Answer 2

Answer:

226.73 kJ

Explanation:

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