Question

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

Answer 1

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 $(\Delta S^o)$.

$\Delta S^o=S_f_{product}-S_f_{reactant}$

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

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

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

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

$\Delta S^o=944.2J/K$

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

$\Delta H^o=H_f_{product}-H_f_{reactant}$

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

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

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

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

$\Delta H^o=-5105kJ$

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:

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

$\Delta G^o=-5386.4kJ$

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