Asked by Ash

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
Answered by DrBob222
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.
Answered by Shauniqua Wiggins
= -1302.5 kJ/mol
Answered by Kahdija
dHrxn = ((2*-393.5)+(-286)) - (227-0)

= -1300 KJ/ mol
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