Question

In a calorimetry experiment the following data were obtained. 

CH₂CH₂(g) + H₂O(ℓ) → C₂H₅OH(ℓ)   ΔH = -288.9kJ

C(s) + O₂(g) → CO₂(g)   ΔH = -395.3kJ

2 H₂(g) + O₂(g) → 2 H₂O(ℓ)   ΔH = -587.4kJ

CH₂CH₂(g) + 3 O₂(g) → 2 CO₂(g) + 2 H₂O(ℓ)   ΔH = -1441.1kJ

Using only this data calculate the molar enthalpy of formation for ethanol, C₂H₅OH(ℓ).

Answer 1

The required equation is

$2C(s)+3H_2(g)+\dfrac{1}{2}O_2(g)\longrightarrow C_2H_5OH(l)\space\space\space\space\space\Delta H=?$

Given equations are

$CH_2CH_2(g)+H_2O(l)\longrightarrow C_2H_5OH(l)\space\space\space\space\space\Delta H_1=-288.9kJ\space(1)$

$C(s)+O_2(g)\rightarrow CO_2(g)\space\space\space\space\Delta H_2=-395.3kJ\space\space(2)$

$2H_2(g)+O_2(g)\rightarrow2H_2O(l)\space\space\space\space\Delta H_3=-587.4kJ\space\space(3)$

$CH_2CH_2(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)\space\space\space\space\Delta H_4=-1441.1kJ\space\space(4)$

Multiplying eq. (2) by 2 and eq.(3) by 3/2 and adding

$2C(s)+3H_2(g)\rightarrow2CO_2(g)+3H_2O(l) \space\space\space\space\space\space\Delta H_5=-1671.1kJ\space\space(5)$

Subtracting eq.(4) from eq.(5) and then adding eq.(1)

$2C(s)+3H_2(g)+\dfrac{1}{2}O_2(g)\rightarrow C_2H_5OH(l)\space\space\space\space\Delta H_f=-518.9kJ/mol$