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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)

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

The standard entropy change of formation, ΔS, 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 = ΔHf - TΔSf

ΔGf = -172 - (298 x (-0.385) = -172 + 114.7 = -57.3 kJ

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.

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.