At T = 25 degrees Celsius, the reaction No2 <-> 2NO +O2 has an equilibrium constant K=5.0*10^-13. Suppose the container is filled with NO2 at an initial pressure of 0.25 atm. Calculate the partial pressure of NO at equilibrium

I assume that K listed is Kp and not Kc. Note that your equation is not balanced.

........2NO2 --> 2NO + O2
I.......0.25......0.....0
C........-2x......2x.....x
E.......0.25-2x....2x......x

Kp = pNO^2*pO2/pNO2^2
Substitute and solve for x and 2x

Well, let's solve this chemical conundrum with a touch of humor!

The equilibrium constant, K, is defined as the ratio of products to reactants' concentrations, but in this case, we'd be dealing with partial pressures. So, NO worries, we can handle that!

To begin, we know that the initial pressure of NO2 is 0.25 atm. Since it's a one-to-one stoichiometric ratio in the balanced equation, the pressure of NO and O2 would be x (since they're both produced in equal amounts).

Using the equilibrium constant expression, we have K = ([NO]^2 * [O2]) / [NO2].

Now, substitute in the given values and solve for x. We obtain:

5.0 * 10^-13 = (x^2 * x) / (0.25).

Rearranging the equation, we have:

x^3 = 5.0 * 10^-13 * 0.25.

Calculating that on my handy-dandy calculator, x ≈ 6.3 * 10^-5 atm.

Thus, the partial pressure of NO at equilibrium is approximately 6.3 * 10^-5 atm.

I hope that helps! If it doesn't, I can always try juggling the numbers and equations, but no promises on that one! 🤡

To solve this problem, we need to use the equilibrium expression and the given information to calculate the partial pressure of NO at equilibrium. Here are the steps:

1. Write the balanced chemical equation for the reaction:
No2 <=> 2NO + O2

2. Define the equilibrium constant expression:
K = [NO]^2 * [O2] / [NO2]

3. From the balance equation, we can see that the stoichiometric coefficient of NO2 is 1, so the equilibrium partial pressure of NO2 is equal to its initial pressure, which is 0.25 atm.

4. Let's assume that at equilibrium, the partial pressure of NO is "x" atm and the partial pressure of O2 is "y" atm.

5. Since the stoichiometric coefficient of NO is 2, the partial pressure of O2 is also "2x" atm.

6. Substitute these values into the equilibrium expression:
K = (x^2)(2x) / 0.25

7. Simplify the expression:
5.0 x 10^(-13) = 2x^3 / 0.25

8. Multiply both sides of the equation by 0.25:
1.25 x 10^(-13) = 2x^3

9. Divide both sides of the equation by 2:
0.625 x 10^(-13) = x^3

10. Take the cube root of both sides of the equation:
x = (0.625 x 10^(-13))^(1/3)

11. Calculate x:
x ≈ 2.69 x 10^(-5) atm

Thus, the partial pressure of NO at equilibrium is approximately 2.69 x 10^(-5) atm.

To calculate the partial pressure of NO at equilibrium, we need to determine what information we have and what equations we can use.

Given:
- Initial pressure of NO2 (P(NO2)) = 0.25 atm,
- Equilibrium constant (K) = 5.0 * 10^-13,
- Reaction: NO2 <-> 2NO + O2.

First, we need to understand that the equilibrium constant, K, is related to the concentrations of the species at equilibrium. The equilibrium constant expression for this reaction is given as:

K = [NO]^2 * [O2] / [NO2],

where [NO], [O2], and [NO2] represent the concentrations of NO, O2, and NO2, respectively.

Since we are given the initial pressure of NO2 (P(NO2)) and not the concentration, we need to use the ideal gas law to convert the pressure to concentration. The ideal gas law is given as:

PV = nRT,

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

To convert the pressure to concentration, we rearrange the ideal gas law equation as:

n/V = P/RT,

where n/V represents the molar concentration (M).

Now, let's do the calculations step by step:

1. Convert the temperature to Kelvin:
T(K) = T(C) + 273.15.
T(K) = 25 + 273.15 = 298.15 K.

2. Convert the pressure of NO2 to concentration (M):
n/V = P(NO2) / RT.
n/V = 0.25 atm / (0.0821 L·atm/(mol·K) * 298.15 K)
= 0.01033 mol/L.

3. As per the balanced equation, the stoichiometry is 1:2:1 for NO2, NO, and O2, respectively. Therefore, at equilibrium, the concentration of NO ([NO]) will be twice the concentration of NO2 ([NO2]). Substituting this information into the equilibrium constant expression:

K = [NO]^2 * [O2] / [NO2],
5.0 * 10^-13 = (2[NO2])^2 * [O2] / [NO2],
5.0 * 10^-13 = 4[NO2]^2 * [O2] / [NO2],

4. Since [O2] is not given, let's assume that the concentration of O2 at equilibrium can be represented as [O2] = x.

5. Substituting the known values and simplifying:

5.0 * 10^-13 = 4(0.01033 mol/L)^2 * x / 0.01033 mol/L,
5.0 * 10^-13 = 4 * 0.01033 mol/L * x,
5.0 * 10^-13 = 0.04132 mol/L * x,
x = (5.0 * 10^-13) / (0.04132 mol/L) = 1.2087 * 10^-11 mol/L.

So, the concentration (or partial pressure) of NO at equilibrium is 1.2087 * 10^-11 mol/L (or atm), respectively.