Answer:
Explanation:
From the given information;
The chemical reaction can be well presented as follows:
⇄ ![\mathtt{3SO_{2(l)}}](https://tex.z-dn.net/?f=%5Cmathtt%7B3SO_%7B2%28l%29%7D%7D)
Now, K is known to be the equilibrium constant and it can be represented in terms of each constituent activity:
i.e
![K = \dfrac{a_{so_3}}{a_{so_2} a_{o_2}^{\frac{1}{2}}}](https://tex.z-dn.net/?f=K%20%3D%20%5Cdfrac%7Ba_%7Bso_3%7D%7D%7Ba_%7Bso_2%7D%20a_%7Bo_2%7D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D)
However, since we are dealing with liquids solutions;
since the activity of
is equivalent to 1
Hence, under standard conditions(i.e at a pressure of 1 bar)
![K = \dfrac{1}{Pso_2Po_2^{1/2}}](https://tex.z-dn.net/?f=K%20%3D%20%5Cdfrac%7B1%7D%7BPso_2Po_2%5E%7B1%2F2%7D%7D)
(b)
From the CRC Handbook, we are meant to determine the value of the Gibb free energy by applying the formula:
![\Delta _{rxn} G^o = \sum \Delta_f \ G^o (products) - \sum \Delta_fG^o (reactants) \\ \\ = (1) (-368 \ kJ/mol) - (\dfrac{1}{2}) (0) - ((1) (-300.13 \ kJ/mol)) \\ \\ = -368 \ kJ/mol + 300.13 \ kJ/mol \\ \\ \simeq -68 \ kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20_%7Brxn%7D%20G%5Eo%20%3D%20%5Csum%20%5CDelta_f%20%5C%20G%5Eo%20%28products%29%20-%20%5Csum%20%5CDelta_fG%5Eo%20%28reactants%29%20%5C%5C%20%5C%5C%20%3D%20%281%29%20%28-368%20%5C%20kJ%2Fmol%29%20-%20%28%5Cdfrac%7B1%7D%7B2%7D%29%20%280%29%20-%20%28%281%29%20%28-300.13%20%5C%20kJ%2Fmol%29%29%20%5C%5C%20%5C%5C%20%3D%20-368%20%5C%20kJ%2Fmol%20%2B%20300.13%20%5C%20kJ%2Fmol%20%5C%5C%20%5C%5C%20%20%5Csimeq%20-68%20%5C%20kJ%2Fmol)
Thus, for this reaction; the Gibbs frree energy = -68 kJ/mol
(c)
Le's recall that:
At equilibrium, the instantaneous free energy is usually zero &
Q(reaction quotient) is equivalent to K(equilibrium constant)
So;
![\mathtt{\Delta _{rxn} G = \Delta _{rxn} G^o + RT In Q}](https://tex.z-dn.net/?f=%5Cmathtt%7B%5CDelta%20_%7Brxn%7D%20G%20%3D%20%5CDelta%20_%7Brxn%7D%20G%5Eo%20%2B%20RT%20In%20Q%7D)
![\mathtt{0- \Delta _{rxn} G^o = RTIn K } \\ \\ \mathtt{ \Delta _{rxn} G^o = -RTIn K } \\ \\ K = e^{\dfrac{\Delta_{rxn} G^o}{RT}} \\ \\ K = e^{^{\dfrac{67900 \ J/mol}{8.314 \ J/mol \times 298 \ K}} }](https://tex.z-dn.net/?f=%5Cmathtt%7B0-%20%5CDelta%20_%7Brxn%7D%20G%5Eo%20%3D%20RTIn%20K%20%7D%20%5C%5C%20%5C%5C%20%5Cmathtt%7B%20%5CDelta%20_%7Brxn%7D%20G%5Eo%20%3D%20-RTIn%20K%20%7D%20%20%5C%5C%20%5C%5C%20%20K%20%3D%20e%5E%7B%5Cdfrac%7B%5CDelta_%7Brxn%7D%20G%5Eo%7D%7BRT%7D%7D%20%5C%5C%20%5C%5C%20%20K%20%3D%20e%5E%7B%5E%7B%5Cdfrac%7B67900%20%5C%20J%2Fmol%7D%7B8.314%20%5C%20J%2Fmol%20%5Ctimes%20298%20%5C%20K%7D%7D%20%7D)
![K =7.98390356\times 10^{11} \\ \\ \mathbf{K = 7.98 \times 10^{11}}](https://tex.z-dn.net/?f=K%20%3D7.98390356%5Ctimes%2010%5E%7B11%7D%20%5C%5C%20%5C%5C%20%20%5Cmathbf%7BK%20%3D%207.98%20%5Ctimes%2010%5E%7B11%7D%7D)
(d)
The direction by which the reaction will proceed can be determined if we can know the value of Q(reaction quotient).
This is because;
If Q < K, then the reaction will proceed in the right direction towards the products.
However, if Q > K , then the reaction goes to the left direction. i.e to the reactants.
So;
![Q= \dfrac{1}{Pso_2Po_2^{1/2}}](https://tex.z-dn.net/?f=Q%3D%20%5Cdfrac%7B1%7D%7BPso_2Po_2%5E%7B1%2F2%7D%7D)
Since we are dealing with liquids;
![Q= \dfrac{1}{1 \times 1^{1/2}}](https://tex.z-dn.net/?f=Q%3D%20%5Cdfrac%7B1%7D%7B1%20%5Ctimes%201%5E%7B1%2F2%7D%7D)
Q = 1
Since Q < K; Then, the reaction proceeds in the right direction.
Hence, SO2 as well O2 will combine to yield SO3, then condensation will take place to form liquid.