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

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

14 days is the half-life of phosphorus-32

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