Question

A.A scientist measures the standard enthalpy change for the following reaction to be -2932.6 kJ :

2C₂H₆(g) + 7 O₂(g)4CO₂(g) + 6 H₂O(g)

Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of CO₂(g) is  kJ/mol.

B.A scientist measures the standard enthalpy change for the following reaction to be -138.9 kJ :

H₂(g) + C₂H₄(g)C₂H₆(g)

Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of C₂H₄(g) is  kJ/mol.

Answer 1

Consider reaction, 2 C₂H₆ (g) + 7 O ₂ (g) $\rightarrow$ 4 CO ₂ (g) + 6 H₂O (g)  

The standard enthalpy change of reaction is given as

r H ⁰ = H ⁰ _f ( products ) -   H ⁰ _f ( reactants )

H ⁰ _f (products) = 4 H ⁰ _f CO _2(g) +6 H ⁰ _f H₂O (g)

H ⁰ _f (products) = 4 H ⁰ _f CO _2(g) + 6 ( -241.82 kJ /mol) =  4 H ⁰ _f CO _2(g) - 1450.092 kJ / mol

H ⁰_f (reactants) =2 H ⁰ _f C₂H₆ (g) + 7 H ⁰ _f O _2(g)

H ⁰_f (reactants) = 2 (-84.0 kJ/ mol ) + 7 ( 0 kJ /mol) = - 168 kJ / mol

From above calculated values , we can write

- 2932.6 kJ / mol = (  4 H ⁰ _f CO _2(g) - 1450.092 kJ / mol ) - ( - 168 kJ / mol )

- 2932.6 kJ / mol = (  4 H ⁰ _f CO _2(g) - 1450.092 kJ / mol ) + 168 kJ / mol  

(  4 H ⁰ _f CO _2(g) - 1450.092 kJ / mol ) = - 2932.6 kJ / mol - 168 kJ / mol = - 3100.6 kJ / mol

(  4 H ⁰ _f CO _2(g) - 1450.092 kJ / mol ) = - 3100.6 kJ / mol

4 H ⁰ _f CO _2(g) = - 3100.6 kJ / mol + 1450.092 kJ / mol = - 1650.51 kJ /mol

H ⁰ _f CO _2(g) = - 1650.51 kJ /mol / 4 = - 412.6 kJ / mol

ANSWER : Standard enthalpy of formation of CO ₂ = - 412.6 kJ / mol

PART 2

Consider reaction, C₂H₄ (g) + H ₂ (g) $\rightarrow$ C ₂H₆ (g)

The standard enthalpy change of reaction is given as

r H ⁰ = H ⁰ _f ( products ) -   H ⁰ _f ( reactants )

H ⁰ _f (products) = H ⁰ _f C ₂H₆ (g) = - 84.0 kJ / mol

H ⁰_f (reactants) = H ⁰ _f  C₂H₄ (g) + H ⁰ _f H _2(g)

H ⁰_f (reactants) = H ⁰ _f  C₂H₄ (g) + 0 kJ / mol = H ⁰ _f  C₂H₄ (g)

From above calculated values , we can write

- 138.9 kJ / mol = ( -84.0 kJ / mol ) - H ⁰ _f  C₂H₄ (g)

H ⁰ _f  C₂H₄ (g) = ( -84.0 kJ / mol ) + 138.9 kJ /mol

H ⁰ _f  C₂H₄ (g) = 54.9 kJ /mol  

ANSWER : Standard enthalpy of formation of C₂H_{4 (g)} = 54.9 kJ / mol