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

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g