Question

If 3.305 g of ethanol C2H5OH(l) is burned completely in a bomb calorimeter at 298.15 K, the heat produced is 97.70 kJ .

A) Calculate ΔH∘combustion for ethanol at 298.15 K

B) Calculate ΔH∘f of ethanol at 298.15 K.

Answer 1

Sol:-

(a) Calculation of delta H⁰_c for Ethanol :-

Given mass of ethanol = 3.305 g

Gram molar mass of ethanol = 46 g/mol

Number of moles of ethanol = Given mass in gram / Gram molar mass = 3.305 g / 46 gmol^-1 = 0.07185 mol

Given Temperature = 298.15 K

and Heat (q) = - 97.70 KJ

delta H⁰_c = q / moles = - 97.70 KJ / 0.07185 mol = -1359.7 KJ/mol

Hence delta H⁰_c of ethanol is -1359.7 KJ/mol

(b). Calculation of delta H⁰_f of Ethanol :-

Combustion reaction of ethanol is :

C₂H₅OH (l) + 3O₂ (g) -------------> 2 CO₂ (g) + 3 H₂O (g) , delta H⁰ = -1359.7 KJ/mol

We know

delta H⁰ = [Sum of enthalpy of formation of products] - [Sum of enthalpy of formation of reactans]

delta H⁰ = [ 2 deltaH_f⁰CO₂(g) + 3deltaH_f⁰ H₂O (g)] - [ deltaH⁰_f C₂H₅OH (l) + 3 deltaH⁰_f O₂ (g)]

- 1359.7 KJ = [ 2(-393.5 KJ) + 3 ( - 285.8 KJ)] - [ deltaH⁰_f C₂H₅OH (g) + 3 (0 KJ) ]

-1359.7 KJ = - 787 KJ - 857.4 KJ - deltaH⁰_f C₂H₅OH (g)

deltaH⁰_f C₂H₅OH (g) =  1359.7 KJ - 1644.4 KJ

deltaH⁰_f C₂H₅OH (g) = - 284.7 KJ