Respuesta :
Part a)
Balanced equation:
2 CH₃OH(l) + 3 O₂(g) → 2 CO₂(g) + 4 H₂O(g)
Part b)
ΔH for the reaction at 25°C
ΔH = Enthalpy change of products - Enthalpy change of reactants
= [(2 mol) * (-393.509 kJ/mol) + (4 mol)*(-241.83 kJ/mol)] - [(2 mol) * (238.4 kJ/mol) + (3 mol) *(0 kJ/mol)] = - 1277.5 kJ
Part c)
ΔS for the reaction at 25°C
ΔS = entropy change in products - entropy change in reactants
= [(2 x 213.7) + (4 x 188.8)] - [(2 x 127.2) + (3 x 205.1)] = 312.9 J K⁻¹
Part d)
ΔG for the reaction at 25°C
ΔG = ΔH - (T * ΔS)
= - 1277.5 x 10³ J - (298 x 312.9 J K⁻¹) = -1370.7 kJ
Balanced equation:
2 CH₃OH(l) + 3 O₂(g) → 2 CO₂(g) + 4 H₂O(g)
Part b)
ΔH for the reaction at 25°C
ΔH = Enthalpy change of products - Enthalpy change of reactants
= [(2 mol) * (-393.509 kJ/mol) + (4 mol)*(-241.83 kJ/mol)] - [(2 mol) * (238.4 kJ/mol) + (3 mol) *(0 kJ/mol)] = - 1277.5 kJ
Part c)
ΔS for the reaction at 25°C
ΔS = entropy change in products - entropy change in reactants
= [(2 x 213.7) + (4 x 188.8)] - [(2 x 127.2) + (3 x 205.1)] = 312.9 J K⁻¹
Part d)
ΔG for the reaction at 25°C
ΔG = ΔH - (T * ΔS)
= - 1277.5 x 10³ J - (298 x 312.9 J K⁻¹) = -1370.7 kJ
The balanced equation for the combustion of methanol is as follows:
2CH3OH + 3O2 → 2CO2 + 4H2O
HOW TO BALANCE CHEMICAL EQUATION?
A balanced equation is an equation whose number of atoms of each element on both sides of the equation is the same.
To balance a chemical reaction, coefficients are used. Coefficients are numbers placed in front of the element/compound.
According to this question, methanol burns in oxygen to form carbon dioxide and water as follows:
2CH3OH + 3O2 → 2CO2 + 4H2O
Learn more about balanced equation at: https://brainly.com/question/8062886