Q466-01
The standard enthalpy change for the combustion of phenol is -3050 kJ mol-1
at 298K. Write an equation for the complete combustion of phenol, C6H5OH(s).
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C6H5OH(s) + 7O2(g) 
6CO2(g) + 3H2O(l) |
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The standard enthalpy changes of formation of carbon dioxide, CO2(g)
and of water, H2O(l) are -394 kJ mol-1 and -286 kJ mol-1
respectively. Calculate the standard enthalpy change of formation of phenol,
C6H5OH(s).
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The enthalpy change of the combustion reaction above = formation
enthalpy of prodcuts - formation enthalpy of reactants
Therefore ΔHc = 6ΔHf(CO2(g)) + 3ΔHf(H2O(l))
- ΔHf(C6H5OH(s))
∴ ΔHf(C6H5OH(s)) = 6ΔHf(CO2(g))
+ 3ΔHf(H2O(l)) - ΔHc
∴ ΔHf(C6H5OH(s)) = [6 x (-394)]
+ [3 x (-286)] - (-3050)
∴ ΔHf(C6H5OH(s)) = -2364
- 858 + 3050 = -172 kJ
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The standard entropy change of formation, ΔS
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of phenol, C6H5OH(s) at 298K is -385 JK-1
mol-1. Calculate the standard free energy of formation, ΔG,
of phenol at 298K.
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The standard free energy change of formation, ΔGf
is given by:
ΔGf
= -172 - (298 x (-0.385) = -172 + 114.7 = -57.3
kJ
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Determine if the reaction is spontaneous at 298K and give a reason.
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| The Gibbs free energy change is negative, so the reaction is
spontaneous at standard temperature. |
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Predict the effect, if any, of an increase in temperature on the spontaneity
of the reaction.
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| The entropy term of the reaction is negative, making the second
part of the right hand side, (-TΔS), positive. Hence, as
the temperature increases Gibbs free energy change becomes less
negative and the reaction becomes less spontaneous. Eventually
a temperature will be reached at which the reaction ceases to
be spontaneous. |
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