Question

In atmospheric chemistry, the following chemical reaction converts SO2, the predominant oxide of sulfur that comes from combustion of S-containing materials, to SO3, which can combine with H2O to make sulfuric acid (and thus contribute to acid rain): a. Write the expression for K for this equilibrium. b. Calculate the value of for this reaction using the values in either the back of your book or the CRC Handbook. c. Calculate the value of K for this equilibrium. d. If 1.00 bar of SO2 and 1.00 bar of O2 are enclosed in a system in the presence of some SO3 liquid, in which direction would the reaction run

Answer 1

Step-by-step explanation:

From the given information;

The chemical reaction can be well presented as follows:

`\mathtt{SO_(2(g)) + (1)/(2)O_(2(g)) }` ⇄
`\mathtt{3SO_(2(l))}`

Now, K is known to be the equilibrium constant and it can be represented in terms of each constituent activity:

i.e

`K = \frac{a_(so_3)}{a_(so_2) a_(o_2)^{(1)/(2)}}`

However, since we are dealing with liquids solutions;

`K = (1)/((Pso_2)/(P^0)\Big ( (Po_2)/(P^0) \Big)^(1/2))` since the activity of
`a_(so_3)` is equivalent to 1

Hence, under standard conditions(i.e at a pressure of 1 bar)

`K = (1)/(Pso_2Po_2^(1/2))`

(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) - ((1)/(2)) (0) - ((1) (-300.13 \ kJ/mol)) \\ \\ = -368 \ kJ/mol + 300.13 \ kJ/mol \\ \\ \simeq -68 \ kJ/mol`

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}`

`\mathtt{0- \Delta _(rxn) G^o = RTIn K } \\ \\ \mathtt{ \Delta _(rxn) G^o = -RTIn K } \\ \\ K = e^{(\Delta_(rxn) G^o)/(RT)} \\ \\ K = e^{^{(67900 \ J/mol)/(8.314 \ J/mol * 298 \ K)} }`

`K =7.98390356* 10^(11) \\ \\ \mathbf{K = 7.98 * 10^(11)}`

(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= (1)/(Pso_2Po_2^(1/2))`

Since we are dealing with liquids;

`Q= (1)/(1 * 1^(1/2))`

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.