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Given the values of δhfo in kj/mol and so in j/mol k given below, calculate the value of δgo in kj for the reaction at 298 k: c6h12o6(s) + 6 o2(g) => 6 co2(g) + 6h2o(g) δhfo (c6h12o6) = -1,277 δhfo (co2) = -396 δhfo (h2o) = -242 so (c6h12o6(s)) = 218 so (o2(g)) = 206 so (co2(g)) = 211 so (h2o(g)) = 18

User Madox
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1 Answer

7 votes

Answer : The value of
\Delta G^o for the reaction is, -5386.4 kJ

Explanation :

The given chemical reaction is:


C_6H_(12)O_6(s)+6O_2(g)\rightarrow 6CO_2(g)+6H_2O(g)

First we have to calculate the entropy of reaction
(\Delta S^o).


\Delta S^o=S_f_(product)-S_f_(reactant)


\Delta S^o=[n_(CO_2(g))* \Delta S^0_((CO_2(g)))+n_(H_2O(g))* \Delta S^0_((H_2O(g)))]-[n_{C_6H_(12)O_6(s)}* \Delta S^0_{(C_6H_(12)O_6(s))}+n_(O_2(g))* \Delta S^0_((O_2(g)))]

where,


\Delta S^o = entropy of reaction = ?

n = number of moles

Now put all the given values in this expression, we get:


\Delta S^o=[6mole* (211J/K.mol)+6mole* (188.7J/K.mol]-[1mole* (218J/K.mol)+6mole* (206J/K.mol]


\Delta S^o=944.2J/K

Now we have to calculate the enthalpy of reaction
(\Delta H^o).


\Delta H^o=H_f_(product)-H_f_(reactant)


\Delta H^o=[n_(CO_2(g))* \Delta H^0_((CO_2(g)))+n_(H_2O(g))* \Delta H^0_((H_2O(g)))]-[n_{C_6H_(12)O_6(s)}* \Delta H^0_{(C_6H_(12)O_6(s))}+n_(O_2(g))* \Delta H^0_((O_2(g)))]

where,


\Delta H^o = enthalpy of reaction = ?

n = number of moles

Now put all the given values in this expression, we get:


\Delta H^o=[6mole* (-396kJ/mol)+6mole* (-242kJ/mol]-[1mole* (-1277kJ/mol)+6mole* (0kJ/mol]


\Delta H^o=-5105kJ

Now we have to calculate the Gibbs free energy.

As we know that,


\Delta G^o=\Delta H^o-T\Delta S^o

where,


\Delta G^o = standard Gibbs free energy = ?


\Delta H^o = standard enthalpy = -5105 kJ


\Delta S^o = standard entropy = 944.2 J/K = 0.9442 kJ/K

T = temperature of reaction = 298 K

Now put all the given values in the above formula, we get:


\Delta G^o=(-5105kJ)-(298K* 0.9442kJ/K)


\Delta G^o=-5386.4kJ

Therefore, the value of
\Delta G^o for the reaction is, -5386.4 kJ

User Vishakha Lall
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