Nitrogen dioxide reacts with SO2 to form NO and SO3:

$$NO2​(g)+SO2​(g)

NO(g)+SO3​(g)

An equilibrium mixture is analyzed at a certain temperature and found to contain [NO2] = 0.100 M,
[SO2] = 0.300 M, [NO] = 2.00 M, and [SO3] = 0.600 M. At the same temperature, extra SO2(g) is added to make a total [SO2] = 0.800 M.

I don't get the $ signs or the $$ signs. Nor do I see a question. I presume that the question is to determine the equilibrium concentrations after the SO2 concentration is changed form 0.300 to 0.800. Here is how you do that. I will delete the $ signs and balance the equation.

.................NO2​(g)+SO2​(g) ==> NO(g)+SO3​(g)
E...............0.100.......0.300...........2.00......0.600
[NO2] = 0.100 M,

First determine Kc. Plug those equilibrium numbers as shown into Kc expression as shown here. Kc = (NO)(SO3)/(NO2)(SO2)
Kc = (2.00)(0.600)/(0.100)(0.300) = 40
Next we redo the equation to show the new concentrations at THE NEW CONDITIONS AS INITIAL and finish the ICE chart.
.................NO2​(g)+SO2​(g) ==> NO(g)+SO3​(g)
I...............0.100.......0.800...........2.00......0.600
C...................-x..........-x................+x...........+x
E...............0.100-x....0.800-x.......2.00+x......0.600+x
Now plug the E line into the Kc expression and solve for x, then evaluate each of the reactants and products concentrations.
Post your work if you get stuck.

Well, looks like it's time for some chemistry talk! You're telling me there's an equilibrium mixture of nitrogen dioxide (NO2), sulfur dioxide (SO2), nitric oxide (NO), and sulfur trioxide (SO3).

Now, let's see what happens when you add some extra SO2 to the mix. Since SO2 is a bit of a troublemaker, let's see how it affects the equilibrium.

When you add more SO2, it will react with the NO according to the equation you mentioned. This will then shift the equilibrium to the right, producing more NO and SO3. So, buckle up, because things are about to get interesting!

The extra SO2 will react with the existing NO and form more NO and SO3. This means the concentration of NO will increase, while the concentration of SO3 will increase as well. On the other hand, the concentrations of NO2 and the original SO2 will remain the same since they are not directly involved in the reaction.

So, in the end, you will end up with even more NO and SO3, but the concentrations of NO2 and SO2 will stay the same. It's like going to a party and ordering more food—some things just never change!

I hope that answers your question. If you have any more chemistry-related queries, I'm here to make you laugh while providing answers!

To determine the effect of adding extra SO2 on the equilibrium mixture, we need to compare the initial concentrations to the new concentrations after adding SO2.

The balanced chemical equation for the reaction is:

NO2(g) + SO2(g) ⇌ NO(g) + SO3(g)

The initial concentrations are:

[NO2] = 0.100 M
[SO2] = 0.300 M
[NO] = 2.00 M
[SO3] = 0.600 M

After adding SO2, the new [SO2] is 0.800 M.

To determine the concentrations of NO and SO3 at equilibrium after adding SO2, we can use an ICE table.

Initial:
[NO2] = 0.100 M
[SO2] = 0.300 M
[NO] = 2.00 M
[SO3] = 0.600 M

Change (due to reaction):
- x (since 1 mole of NO2 produces 1 mole of NO)
- x (since 1 mole of NO2 produces 1 mole of NO)
+ x (since 1 mole of NO2 produces 1 mole of NO)
+ x (since 1 mole of NO2 produces 1 mole of SO3)

Equilibrium:
[NO2] = 0.100 M - x
[SO2] = 0.800 M
[NO] = 2.00 M + x
[SO3] = 0.600 M + x

At equilibrium, the concentrations of reactants and products remain constant, so the changes in concentration are equal to zero.

[NO2] = 0.100 M - x = 0 (since [NO2] does not change at equilibrium)
0.100 M - x = 0
x = 0.100 M

[NO] = 2.00 M + x = 2.00 M + 0.100 M = 2.10 M
[SO3] = 0.600 M + x = 0.600 M + 0.100 M = 0.700 M

Therefore, after adding extra SO2, the new equilibrium concentrations are:

[NO2] = 0.100 M
[SO2] = 0.800 M
[NO] = 2.10 M
[SO3] = 0.700 M

To find the new concentrations of NO, NO2, and SO3 after adding extra SO2, we need to determine the reaction quotient Q and compare it to the equilibrium constant K.

1. Write the balanced chemical equation:

NO2(g) + SO2(g) ⇌ NO(g) + SO3(g)

2. Determine the initial concentrations:

[NO2] = 0.100 M
[SO2] = 0.300 M
[NO] = 2.00 M
[SO3] = 0.600 M

3. Calculate the initial reaction quotient Q for the given concentrations:

Q = [NO]^a * [SO3]^b / ([NO2]^c * [SO2]^d)

Since the coefficients in the balanced equation are both 1, the exponents a, b, c, and d are all 1.

Q = [NO]^1 * [SO3]^1 / ([NO2]^1 * [SO2]^1)

Q = [NO] * [SO3] / ([NO2] * [SO2])

Substituting the given values into the equation, we get:

Q = 2.00 * 0.600 / (0.100 * 0.300)

Q = 12 / 0.030

Q = 400

4. Compare Q to the equilibrium constant K:

If Q < K, the reaction will proceed in the forward direction to reach equilibrium.
If Q > K, the reaction will shift in the reverse direction to reach equilibrium.
If Q = K, the system is already at equilibrium.

5. If Q < K (400 < K), then adding extra SO2 will shift the reaction to the forward direction, resulting in an increase in the concentrations of NO and SO3, and a decrease in the concentrations of NO2 and SO2.

The new concentrations after adding extra SO2 are:
[NO2] = 0.100 M - x, where x is the change in concentration
[SO2] = 0.800 M - x
[NO] = 2.00 M + x
[SO3] = 0.600 M + x

However, we need to determine the value of x to calculate the actual new concentrations.

6. Use the stoichiometric ratios from the balanced equation to relate the changes in concentrations:

From the balanced equation, the stoichiometric ratio of NO2 to SO2 is 1:1, and the stoichiometric ratio of NO to SO3 is 1:1.

Therefore, x = [NO2] = [SO2]

7. Substitute x back into the expressions for the new concentrations:

[NO2] = 0.100 M - x = 0.100 M - [NO2]
[SO2] = 0.800 M - x = 0.800 M - [SO2]
[NO] = 2.00 M + x = 2.00 M + [NO2]
[SO3] = 0.600 M + x = 0.600 M + [SO2]

Simplifying the equations:

[NO2] = 0.100 M - [NO2]
[SO2] = 0.800 M - [SO2]
[NO] = 2.00 M + [NO2]
[SO3] = 0.600 M + [SO2]

Rearranging the equations:

2[NO2] = 0.100 M
2[SO2] = 0.800 M
[NO] = 2.00 M + [NO2]
[SO3] = 0.600 M + [SO2]

Solving the equations, we find:

[NO2] = 0.050 M
[SO2] = 0.400 M
[NO] = 2.05 M
[SO3] = 1.00 M

So, after adding extra SO2 to make a total [SO2] = 0.800 M, the new equilibrium concentrations are [NO2] = 0.050 M, [SO2] = 0.400 M, [NO] = 2.05 M, and [SO3] = 1.00 M.