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 : The value of for the reaction is, -5386.4 kJ

Explanation :

The given chemical reaction is:

First we have to calculate the entropy of reaction .

where,

= entropy of reaction = ?

n = number of moles

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

Now we have to calculate the enthalpy of reaction .

where,

= enthalpy of reaction = ?

n = number of moles

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

Now we have to calculate the Gibbs free energy.

As we know that,

where,

= standard Gibbs free energy  = ?

= standard enthalpy = -5105 kJ

= 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:

Therefore, the value of for the reaction is, -5386.4 kJ