At a certain temperature, the equilibrium constant for the following chemical equation is 2.00. At this temperature, calculate the number of moles of NO2(g) that must be added to 2.64 mol of SO2(g) in order to form 1.10 mol of SO3(g) at equilibrium.

SO2(g) + NO2(g) <--->SO3(g) + NO(g)

Do a RICE table

R: SO2(g) + NO2(g) <--> SO3(g) + NO(g)
I: 2.64 z 0 0
C: -X - x +x +x
E: 2.64-x z-x x x

the equation tells you what x is
Since 1.10 mol of SO3(g) is at equilibrium x must be 1.10

Revised E:
2.64 - 1.10, z - 1.10, 1.10, 1.10

[1.10][1.10]
------------ = 2
[1.54][z-1.10]

Solve for z.

Why did the chemist break up with his SO2(g) molecule?

Because it wasn't giving him enough SO3(g) in return!

But let's get serious for a moment and do some calculations.

We're given that the equilibrium constant (K) for this reaction is 2.00. At equilibrium, the ratio of products to reactants can be determined using the stoichiometry of the reaction.

From the balanced equation, we can see that for every 1 mole of SO2(g) that reacts, 1 mole of SO3(g) is produced. And for every 1 mole of SO2(g) that reacts, 1 mole of NO2(g) reacts as well.

So, if we start with 2.64 mol of SO2(g) and want to form 1.10 mol of SO3(g), we can determine the required amount of NO2(g) using the stoichiometry.

Since the ratio of SO2(g) to SO3(g) is 1:1, we can say that if we form 1.10 mol of SO3(g), we need 1.10 mol of SO2(g). And since the ratio of SO2(g) to NO2(g) is also 1:1, we would need 1.10 mol of NO2(g) as well.

Therefore, you would need to add 1.10 mol of NO2(g) to 2.64 mol of SO2(g) in order to form 1.10 mol of SO3(g) at equilibrium.

To solve this problem, we need to use the concept of stoichiometry and the equilibrium constant expression.

First, let's determine the initial moles of each substance and the change in moles.

Initial moles:
SO2(g) = 2.64 mol
NO2(g) = 0 mol
SO3(g) = 0 mol
NO(g) = 0 mol

Change in moles:
SO2(g) = -x mol
NO2(g) = +x mol
SO3(g) = +x mol
NO(g) = +x mol

At equilibrium, the concentrations of the reactants and products can be calculated using the equation:

Kc = ([SO3(g)] * [NO(g)]) / ([SO2(g)] * [NO2(g)])

Given that the equilibrium constant (Kc) is 2.00 and the concentrations at equilibrium are:
[SO3(g)] = 1.10 mol/L
[SO2(g)] = 2.64 mol/L
[NO2(g)] = x mol/L

We can substitute these values into the equation and solve for x:
2.00 = ((1.10)(x)) / ((2.64)(x))
2.00 = 1.10 / 2.64
2.00 = 0.416

To simplify, we can multiply both sides by 2.64 to get:
2.64 = 0.416x

Now, divide both sides by 0.416 to solve for x:
x = 2.64 / 0.416

x ≈ 6.35

Therefore, approximately 6.35 moles of NO2(g) must be added to 2.64 moles of SO2(g) to form 1.10 moles of SO3(g) at equilibrium.

To solve this problem, we need to use the concept of stoichiometry and the equilibrium constant expression.

First, let's write down the balanced equation:

SO2(g) + NO2(g) <--> SO3(g) + NO(g)

The equilibrium constant expression for this equation is:

Kc = [SO3][NO] / [SO2][NO2]

Given that the equilibrium constant (Kc) is 2.00, we can set up the following equation:

2.00 = ([SO3]eq * [NO]eq) / ([SO2]eq * [NO2]eq)

Now, let's calculate the equilibrium concentrations using the given information:

[SO2]eq = 2.64 mol
[SO3]eq = 1.10 mol

We need to find the number of moles of NO2(g) that must be added, so let's assume that the initial amount of NO2(g) is x mol. Therefore, at equilibrium, the concentrations would be:

[NO2]eq = x mol
[NO]eq = x mol

Substituting these values into the equilibrium constant expression, we get:

2.00 = (1.10 * x) / (2.64 * x)

Cross-multiplying and simplifying:

2.00 * 2.64 * x = 1.10 * x
5.28 * x = 1.10 * x
5.28 = 1.10

Dividing both sides by 1.10:

x = 4.8

Therefore, 4.8 moles of NO2(g) must be added to 2.64 moles of SO2(g) in order to form 1.10 moles of SO3(g) at equilibrium.