if possible please explain ! I am still iffy about this topic. Thanks 6) Conside

Question

if possible please explain ! I am still iffy about this topic. Thanks

6) Consider the following reaction:

SO2Cl2 (g) SO2 (g) + Cl2 (g)

A. What is the oxidation state of S?
S in SO2Cl2: _____________

Kc =2.99 X 107 (4 points)

S in SO2: _____________

Is SO2Cl2 a reducing agent or an oxidizing agent? Circle your answer. (2 points) Reducing agent Oxidizing agent

If initial concentration of SO2Cl2 is 0.195 M, determine the relationship between Qc and Kc for the given reaction. Circle your answer.

(2 points)

Qc > Kc Qc < Kc

Calculate the equilibrium concentration of Cl2. Also, check for 5% rule

[SO2Cl2]

[SO2]

[Cl2]

Initial

Change

Equilibrium

______________ M

Check assumption: ?? / 0.195 X 100% =

E. Formation of SO2 is an exothermic reaction. Circle the direction in which the

reaction will proceed when the following changes are made:

Temperature is increased Left Right No Change

Cl2 is removed Left Right No Change

Volume is increased Left Right No Change

[SO2Cl2]

[SO2]

[Cl2]

Initial

Change

Equilibrium

Explanation / Answer

SO2Cl2 (g) SO2 (g) + Cl2 (g)

The oxidation state is calculated assuming some oxidation state for the elements here. In the case of SO2, we have that O is usually working with 2- oxidation state, so:

2 atoms of oxygen * (-2) + X * atom of S = 0
X = +4 This is the oxidation state in SO2

In SO2Cl2 it would be:

2*(-2) + X + 2(-1) = 0
X = +6

So we have that S pass from 6+ to 4+, this means that is reducting (gaining electrons) while Cl is oxidizing (lose electrons).

Now for equilibrium concentration:

SO2Cl2 (g) SO2 (g) + Cl2 (g) Kc = 2.99x10-7
i: 0.195 0 0
c: -x +x +x
e: 0.195-x x x

2.99x10-7 = x2 / 0.195-x --> Kc is really low value, so we can assume that X will be a small value too, and we can approximate 0.195-x to 0.195 only, and avoid a quadratic formula so:
2.99x10-7 * 0.195 = x2
x = [Cl2] = 2.41x10-4 M

the check assumption would be:
% = 0.195-2.41x10-4 / 0.195 * 100 = 99.88 %. The assumption was correct.

If this is an exothermic reaction an increase in temperature will be favoring the reactants, so, the reaction will go left.
If Cl2 is removed, then the equilibrium will shift to the side where the loss is. will shift to the products (right)
If volume is increased there would be no change.

Hope this helps

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