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

Question

Chemistry

Conley

9 August, 20:26

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

+2

Answers (1)

Know the Answer?

Aspen Oconnor · Answer 1

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.