Question

The standard enthalpies of formation, at 25.00 ^oC, of methanol (CH₄O(l)), water (H₂O(l)), and carbon dioxide (CO₂(g)) are, respectively, -238.7 kJ/mol, -285.8 kJ/mol, and -393.5 kJ/mol. Calculate the change in the entropy of the surroundings (in J/K) upon the combustion of 16.9 g of methanol under a constant pressure of 1.000 atm and a temperature of 25.00 ^oC. N.B. combustion is the reaction of this substance with molecular oxygen to produce water and carbon dioxide.

Answer 1

The balanced chemical equation for the reaction of combustion of methonal is

2CH4O(l) + 3O2(g) ------> 2CO2(g) + 4H2O(l)   

standard enthalphy of reaction(∆H°) is calculated by standard enthalphy of formation(∆H°_f) by the following equation

∆H° = n∆H°_f(products) - n∆H°_f(reactants)

= ( 2mol × ∆H°_f(CO2(g)) + 4mol × ∆H°_f(H2O(l))) - ( 2mol × ∆H°_f(CH4O(l)))

= ( 2mol × -393.5kJ/mol + 4mol × -285.8kJ/mol) - ( 2mol × -238.7kJ/mol)

= -1930.2kJ+ 238.7kJ

= -1691.5kJ

given moles of CH4O = 16.9g/ 32.05g/mol = 0.5273mol

∆H° for combustion of 0.5273moles of CH4O = (-1691.5kJ/2mol)× 0.5273mol = -445.96kJ

∆S_surr = -∆H/T

∆S_surr = - ( -445.96kJ/298.15K)

∆S_surr =  1.496kJ{K

∆S_surr = 1496 J/K