SO this is my first time doing this... lol i need help on an AP chemistry question for equilibrium.

A 0.500 L tank contains 3.00 g of NO(g) at 750. K. The equilibrium constant for the reaction
below at this temperature is 3.4 x 10
-3
2NO(g) ⇌ N2(g) + O2(g)

(a) write the equilibrium constant expression for this reaction:

(b) calculate Kc for the reaction 2N2(g) + 2O2(g) ⇌ 4NO(g)

(c) determine the equilibrium concentrations of all species

moles NO = 3.00g/molar mass = 3.00/30 = 0.1 mol.

a. I assume you can do this. It's just products/reactants with coefficients becoming exponents.

b. This reaction is doubled and reversed. Take the reciprocal of the original Kc to take care of the reversed part and square everything to take care of the doubled part.

c. I assume this is for the original K from part a.
.........2NO ==> N2 + O2
initial.0.1......0....0
change..-2x......x.....x.
equil...0.1-2x....x.....x

Substitute the equilibrium values from the ICE chart into Kc expression (the one you wrote in part a) and solve for x and 0.1-2x.

That's wrong

To solve this AP chemistry question on equilibrium, let's break it down step by step.

(a) Writing the equilibrium constant expression:
The general form of an equilibrium constant expression is given by:
Kc = [products] / [reactants]

In this case, the reaction is:
2NO(g) ⇌ N2(g) + O2(g)

So, the equilibrium constant expression would be:
Kc = [N2] * [O2] / [NO]^2

(b) Calculating Kc for the reaction:
The given reaction is:
2N2(g) + 2O2(g) ⇌ 4NO(g)

Since the given reaction is the reverse of the reaction in part (a), we can use the reciprocal of the equilibrium constant from part (a) to calculate Kc for this reaction.

Kc = 1 / (Kc of part (a))
Kc = 1 / (3.4 x 10^-3)

(c) Determining the equilibrium concentrations of all species:
To determine the equilibrium concentrations, we'll need to use the given information along with the balanced equation and the value of Kc.

Given information:
Volume of the tank = 0.500 L
Mass of NO(g) = 3.00 g
Temperature = 750 K
Equilibrium constant (Kc) = 3.4 x 10^-3

First, we need to calculate the initial molar concentration of NO(g):
Use the equation: moles = mass / molar mass

Molar mass of NO(g) = 30.01 g/mol
Moles of NO(g) = 3.00 g / 30.01 g/mol

Next, we can assume that initially, there is no N2(g) and O2(g), so their concentrations are zero.

Now, we can set up the ICE table to determine the equilibrium concentrations using the given information and the value of Kc.

Initial:
NO(g): [NO] = moles of NO(g) / volume of the tank
N2(g): [N2] = 0
O2(g): [O2] = 0

Change:
NO(g): -2x
N2(g): x
O2(g): x

Equilibrium:
NO(g): [NO] - 2x
N2(g): x
O2(g): x

At equilibrium, the concentration of NO(g) would be equal to the initial concentration minus 2 times the change (2x). The concentrations of N2(g) and O2(g) would be equal to x.

To solve for x, you can substitute the equilibrium concentrations into the equilibrium constant expression:

3.4 x 10^-3 = [N2] * [O2] / ([NO] - 2x)^2

Now, you can solve this equation for x using algebraic methods or numerical methods, such as trial and error.

Once you find the value of x, you can substitute it back into the equilibrium concentrations to find the numerical values of [NO], [N2], and [O2].

I hope this step-by-step explanation helps you understand how to approach and solve this AP chemistry equilibrium question.