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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

Sagot :

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Answer:

Explanation:

From the given information;

The chemical reaction can be well presented as follows:

[tex]\mathtt{SO_{2(g)} + \dfrac{1}{2}O_{2(g)} }[/tex]  ⇄ [tex]\mathtt{3SO_{2(l)}}[/tex]

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

i.e

[tex]K = \dfrac{a_{so_3}}{a_{so_2} a_{o_2}^{\frac{1}{2}}}[/tex]

However, since we are dealing with liquids solutions;

[tex]K = \dfrac{1}{\dfrac{Pso_2}{P^0}\Big ( \dfrac{Po_2}{P^0} \Big)^{1/2}}[/tex]   since the activity of [tex]a_{so_3}[/tex] is equivalent to 1

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

[tex]K = \dfrac{1}{Pso_2Po_2^{1/2}}[/tex]

(b)

From the CRC Handbook, we are meant to determine the value of the Gibb free energy by applying the formula:

[tex]\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[/tex]

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;

[tex]\mathtt{\Delta _{rxn} G = \Delta _{rxn} G^o + RT In Q}[/tex]

[tex]\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}} }[/tex]

[tex]K =7.98390356\times 10^{11} \\ \\ \mathbf{K = 7.98 \times 10^{11}}[/tex]

(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;

[tex]Q= \dfrac{1}{Pso_2Po_2^{1/2}}[/tex]

Since we are dealing with liquids;

[tex]Q= \dfrac{1}{1 \times 1^{1/2}}[/tex]

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.

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