Calculate the standard enthalpy change and the standard entropy change for the combustion of 1 mole of acetylene (C2H2) at 25oC under 1 atm. pressure.
C2H2(g) + 2.5 O2(g) 2 CO2(g) + H2O(l)
SΘ [C2H2(g)] = 200.94 J/(K.mol) SΘ [CO2(g)] = 213.74 J/(K.mol)
SΘ [O2(g)] = 205.14 J/(K.mol) SΘ [H2O(l)] = 69.91 J/(K.mol)
ΔHϴf [CO2(g)] = -393.5 kJ/mol; ΔHϴf [H2O(l)] = -286 kJ/mol; ΔHϴf [C2H2(g)] = 227 kJ/mol
dHrxn = ((2*-393.5)+(-286)) - (227-0)
= -1300 KJ/ mol
C2H2(g) + 2.5 O2(g) 2 CO2(g) + H2O(l)
dHrxn = (n*dHo products) - (n*dHo reactants)
Substitute the numbers and solve for dH rxn
Post your work if you get stuck.
Well, let's get into some chemistry calculations! To calculate the standard enthalpy change (ΔHϴ) for the combustion of acetylene, we need to use the enthalpies of formation for the reactants and products.
ΔHϴ = (sum of products enthalpies) - (sum of reactants enthalpies)
ΔHϴ = (2 mol CO2 x -393.5 kJ/mol) + (1 mol H2O x -286 kJ/mol) - (1 mol C2H2 x 227 kJ/mol)
ΔHϴ = -787 kJ/mol
So, the standard enthalpy change for the combustion of 1 mole of acetylene is -787 kJ/mol.
To calculate the standard entropy change (ΔSϴ), we can use the standard entropies of the reactants and products.
ΔSϴ = (sum of products entropies) - (sum of reactants entropies)
ΔSϴ = 2 mol CO2 x (213.74 J/(K.mol)) + 1 mol H2O x (69.91 J/(K.mol)) - 1 mol C2H2 x (200.94 J/(K.mol)) - 2.5 mol O2 x (205.14 J/(K.mol))
ΔSϴ = -35.05 J/(K.mol)
So, the standard entropy change for the combustion of 1 mole of acetylene is -35.05 J/(K.mol).
Now, for the punchline:
Why don't scientists trust atoms?
Because they make up everything!
To calculate the standard enthalpy change (ΔH°) and the standard entropy change (ΔS°) for the combustion of 1 mole of acetylene (C2H2) at 25°C under 1 atm pressure, you first need to determine the stoichiometric coefficients of the balanced equation.
The balanced equation for the combustion of 1 mole of acetylene (C2H2) is:
C2H2(g) + 2.5 O2(g) → 2 CO2(g) + H2O(l)
Now, we can calculate the ΔH° and ΔS° using the equation:
ΔH° = ΣΔH°(products) - ΣΔH°(reactants)
ΔS° = ΣS°(products) - ΣS°(reactants)
Let's calculate ΔH° first:
ΔH° = 2ΔH°(CO2) + ΔH°(H2O) - ΔH°(C2H2) - 2.5ΔH°(O2)
Substituting the given values:
ΔH° = 2(-393.5 kJ/mol) + (-286 kJ/mol) - (227 kJ/mol) - 2.5(0 kJ/mol)
Simplifying the equation:
ΔH° = -786 kJ/mol - 286 kJ/mol - 227 kJ/mol
ΔH° = -1299 kJ/mol
The standard enthalpy change for the combustion of 1 mole of acetylene is ΔH° = -1299 kJ/mol.
Now, let's calculate ΔS°:
ΔS° = 2S°(CO2) + S°(H2O) - S°(C2H2) - 2.5S°(O2)
Substituting the given values:
ΔS° = 2(213.74 J/(K.mol)) + (69.91 J/(K.mol)) - (200.94 J/(K.mol)) - 2.5(205.14 J/(K.mol))
Simplifying the equation:
ΔS° = 427.48 J/(K.mol) + 69.91 J/(K.mol) - 200.94 J/(K.mol) - 512.85 J/(K.mol)
ΔS° = -215.40 J/(K.mol)
The standard entropy change for the combustion of 1 mole of acetylene is ΔS° = -215.40 J/(K.mol).
Therefore, the standard enthalpy change (ΔH°) is -1299 kJ/mol and the standard entropy change (ΔS°) is -215.40 J/(K.mol) for the combustion of 1 mole of acetylene at 25°C under 1 atm pressure.