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Energetics, Hess' law
Help!
Just need clarification on how you would solve this:
calculate standard enthalpy of formation of ethanoic acid given:
2C(s) + 2H2(g) + O2(g) -----> CH3COOH(l)
when enthalpy of combustion of:
C(s) = -394 kJmol-1
H2(g) = -286 kJmol-1
CH3COOH(l) = -1444 kJmol -1
Im unsure of what I do with the numbers, what would be the alternate route? and how would I go about making a hess' cycle??
much appreciated
Just need clarification on how you would solve this:
calculate standard enthalpy of formation of ethanoic acid given:
2C(s) + 2H2(g) + O2(g) -----> CH3COOH(l)
when enthalpy of combustion of:
C(s) = -394 kJmol-1
H2(g) = -286 kJmol-1
CH3COOH(l) = -1444 kJmol -1
Im unsure of what I do with the numbers, what would be the alternate route? and how would I go about making a hess' cycle??
much appreciated
Enthalpy of formation = (enthalpy of products) - (enthalpy of reactants)
so, enthalpy of formation = (-1444) - (2 x -394 + 2 x -286)
= -84 kJmol -1
so, enthalpy of formation = (-1444) - (2 x -394 + 2 x -286)
= -84 kJmol -1
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