Question

The combustion of one mole of butane takes place at 298 k and 1 atm .After combustion carbondioxide (g) and H2O(water (l)) are produced and 2878.7 kJ/mol heat is liberated . Compute the standard enthalpy of formation of butane. Given that standard enthalpy of formation of carbondioxide (g) and H2O(l) are -393.5 kJ/mol and -285.5kJ/mol. respectively.

Answer 1

GIVEN DATA:

(1)C₄H₁₀+ 13/2O₂ -----> 4CO₂ + 5H₂0 ​ΔH_combustion =2878.7kJ/mole
(2) C_(s) + O_2(g) ==> CO_{2 (g)} ΔH₁ = -393.5 kJmol^-1
(3) H_{2 (g)} + ¹/₂O_{2 (g)} ==> H₂O_(l) ΔH ₂ = -285.8 kJmol^-1

(4) 4 C _(s) + 5H_2(g) ==> C₄H_{10 (g)} ΔH₃ = ????

​
Balancing the equation and then solving as follows :

a) 4C_(s) + 4O_{2 (g)} ==> 4CO_{2 (g)} 4ΔH1 = -4*393.5 kJmol^-1​
b) 5 H_{2 (g)} + 5/₂O_{2 (g)} ==> 5H₂O_(l) 5 ΔH 2 = -5*285.8 kJmol-1

c ) C₄H₁₀ -------> 4C +5H2 ​-ΔH3=???

​adding equation a,b,c we get
C₄H₁₀ + 13/2O₂ -----> 4CO₂ + 5H₂0


4ΔH₁ +​ 5ΔH₂ -ΔH₃ =​ΔH_comustion


calculating we get
​ΔH₃ = -5881.7 kJ/mole