The enthalpy of combustion of acetylene C2H2 is described by C2H2 (g) + (5/2) O2 (g) →2CO2 (g) + H2O (l),  H rxn = - 1299 kJ/mol. Calculate the enthalpy of formation of acetylene, given the following enthalpies of formation  H f[CO2 (g) ] = - 393.5 kJ/mol  H f[H2O (l) ] = - 285.8 kJ/mo

Question

Hana Roman

Chemistry

22 October, 01:20

The enthalpy of combustion of acetylene C2H2 is described by C2H2 (g) + (5/2) O2 (g) →2CO2 (g) + H2O (l),  H rxn = - 1299 kJ/mol. Calculate the enthalpy of formation of acetylene, given the following enthalpies of formation  H f[CO2 (g) ] = - 393.5 kJ/mol  H f[H2O (l) ] = - 285.8 kJ/mo

+2

Answers (2)

Know the Answer?

Bradley Swanson · Answer 1

The enthalpy of formation of acetylene is 226.2 kJ/mol

Explanation:

Step 1: Data given

C2H2 (g) + (5/2) O2 (g) → 2CO2 (g) + H2O (l) Heat of Reaction (Rxn) = - 1299kJ/mol

Standard formation [CO2 (g) ] = - 393.5 kJ/mol

Standard formation [H2O (l) ] = - 285.8 kj/mol

Step 2: The balanced equation

The formation of acetylene is:

2C (s) + H2 (g) → C2H2 (g)

Step 3: Calculate the enthalpy of formation of acetylene

It doesn't matter if the process will happen in 1 step or in more steps. What matters is the final result. This is Hess' law of heat summation.

To have the reaction of the formation of acetylene we have to take:

⇒ the reverse equation of the combustion of acetylene

2CO2 (g) + H2O (l) → C2H2 (g) + (5/2) O2 (g)

⇒ The equation of formation of CO2 (multiplied by 2)

2C (s) + 2O2 (g) → 2CO2 (g)

⇒ the equation of formation of H2O

H2 (g) + 1/2 O2 (g) → H2O (l)

2CO2 (g) + H2O (l) + 2C (s) + 2O2 (g) + H2 (g) + 1/2 O2 (g → C2H2 (g) + (5/2) O2 (g) + 2CO2 (g) + H2O (l)

Final reaction = 2C (s) + H2 (g) → C2H2 (g)

Calculate the enthalpy of formation of acetylene =

ΔHf = 1299 kJ/mol + (2*-393.5) kJ/mol + (-285.8) kJ/mol

ΔHf = 226.2 kJ/mol

The enthalpy of formation of acetylene is 226.2 kJ/mol

Wayne · Answer 2

226.2 kJ/mol

Explanation:

Let's consider the following thermochemical equation for the combustion of acetylene.

C₂H₂ (g) + (5/2) O₂ (g) → 2 CO₂ (g) + H₂O (l), ΔH°rxn = - 1299 kJ/mol.

We can also calculate the enthalpy of the reaction per mole of acetylene from the enthalpies of formation.

ΔH°rxn = 2 mol * ΔH°f (CO₂ (g)) + 1 mol * ΔH°f (H₂O (l)) - 1 mol * ΔH°f (C₂H₂ (g)) - 1 mol * ΔH°f (O₂ (g))

1 mol * ΔH°f (C₂H₂ (g)) = 2 mol * ΔH°f (CO₂ (g)) + 1 mol * ΔH°f (H₂O (l)) - ΔH°rxn - 1 mol * ΔH°f (O₂ (g))

1 mol * ΔH°f (C₂H₂ (g)) = 2 mol * (-393.5 kJ/mol) + 1 mol * (-285.8 kJ/mol) - (-1299 kJ) - 1 mol * (0 kJ/mol)

ΔH°f (C₂H₂ (g)) = 226.2 kJ/mol