At 200 °C, Kc = 1.4 x 10-10 for the reaction

N2O(g) + NO2(g) 3NO(g)
If 0.174 mol of N2O and 0.337 mol NO2 are placed in a 4.00 L container, what would the NO concentration be if this equilibrium were established?

To find the NO concentration at equilibrium, we need to use the given equilibrium constant (Kc) and calculate the concentrations of reactants and products.

Step 1: Write the balanced equation for the reaction:
N2O(g) + NO2(g) ⇌ 3NO(g)

Step 2: Define the initial concentrations of N2O and NO2:
[N2O] = 0.174 mol / 4.00 L = 0.0435 M
[NO2] = 0.337 mol / 4.00 L = 0.0843 M

Step 3: Determine the change in concentration for the reaction. Since the stoichiometry of the reaction is 1:1 for N2O and NO2, and 3:1 for NO, the change will be as follows:
[N2O] = -x
[NO2] = -x
[NO] = +3x

Step 4: Express the equilibrium concentrations in terms of the initial concentrations and the change:
[N2O] = 0.0435 - x
[NO2] = 0.0843 - x
[NO] = 3x

Step 5: Substitute the equilibrium concentrations into the equilibrium expression, using the given equilibrium constant (Kc):
Kc = [NO]^3 / ([N2O] * [NO2])
1.4 x 10^-10 = (3x)^3 / ((0.0435 - x) * (0.0843 - x))

Step 6: Solve the equation for x. This step involves solving a cubic equation, which can be challenging. However, we can estimate the value of x by assuming that x is small compared to the initial concentrations of N2O and NO2. Therefore, we can approximate 0.0435 - x as 0.0435 and 0.0843 - x as 0.0843:
1.4 x 10^-10 ≈ (3x)^3 / (0.0435 * 0.0843)

Step 7: Solve the equation for x. Rearrange the equation to solve for x:
(3x)^3 = 1.4 x 10^-10 * (0.0435 * 0.0843)
x ≈ ∛[(1.4 x 10^-10 * 0.0435 * 0.0843) / 27]

Using a calculator, calculate:
x ≈ 1.164 x 10^-5 M

Step 8: Calculate the NO concentration at equilibrium:
[NO] = 3x ≈ 3 * 1.164 x 10^-5 ≈ 3.492 x 10^-5 M

Therefore, the concentration of NO at equilibrium would be approximately 3.492 x 10^-5 M.

You need to find the arrow keys on your computer and use them.

(N2O) = 0.174mol/4.00L = estimated 0.04
(NO2) = 0.337 mol/4.00L = estd 0.08
...........N2O(g) + NO2(g)==> 3NO(g)
I.........0.04.....0.08........0
C.........-x........-x.........3x
E.........0.04-x...0.08-x......3x

Substitute the E line into Kc expression and solve for x.
Then (NO) = 3x