Question

a. Define standard enthalpy change of formation.

b. Calculate the standard enthalpy change of formation of methane from the following standard enthalpy changes of combustion:
carbon = –394 kJ mol–1
hydrogen = –286 kJ mol–1
methane = –891 kJ mol–1

c. Calculate the standard enthalpy change of combustion of methane using the following bond energies:
E(C — H) = +412 kJ mol–1
E(O — O) = +496 kJ mol–1
E(C — O) = +805 kJ mol–1
E(O — H) = +463 kJ mol–1

Answer 1

A ) The standard enthalpy of formation of a substance is the enthalpy change that occurs when 1 mole of the substance is formed from its constituent elements in their standard states. A pure element in its standard state has a standard enthalpy of formation of zero.

 

B ) C - 394 + (- 286 × 2) - (- 726)

C(s) + O2 (g) → CO 2(g)

ΔH c =−394kJmol −1 .....(1)

 

H2(g) +1/2 O2(g) →H 2O(l)

ΔH c =286kJ /mol

 

2×[H 2(g) + 1/2 O2O (g) →H2O (l) ;ΔH c =−286kJmol −1]

2H 2(gg +O 2 (gg →2H2O (l)

 ;ΔH c = −572kJmol −1 .....(2)

 

CH3OH (l) + 3/2 O2 (g) →CO 2 (g) +2H2O(l)

 ;ΔH c =−726kJmol −1

 

CO2 (g) +2H2O (l). → CH3O(l) - 3/2O2 (g)

 ;ΔHc =726kJmol −1 .....(3)

Adding equation (1),(2)&(3), we get

 

C (s)+2H2 (g)+1/2 O2 (g) →CH3OH (l)

 ;ΔH c =−394+(−286×2)−(−726)

 

C) ∆nreaction =Σ Breactant - ΣB.Eproduct

=( 4× B.E.C-n+2×B.E0 ) = 0

= (1640+992) - (1610-1840)

= 2632 - 3450

= -818KJ/mol

 

 

∆nreaction =Σ Breactant - ΣB.Eproduct

=( 4× B.E.C-n+2×B.E0 ) = 0

= (1640+992) - (1610-1840)

= 2632 - 3450

= -818KJ/mol