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
To calculate the partial pressure of NO at equilibrium, we need to consider the stoichiometry and the equilibrium constant of the reaction. The equilibrium constant, K, is given as the ratio of the products to reactants, each raised to the power of their stoichiometric coefficient.
For the reaction: NO2 <-> 2NO + O2
The stoichiometric coefficients are 1 for NO2, 2 for NO, and 1 for O2.
The equilibrium constant expression for this reaction is:
K = ([NO]^2 * [O2]) / [NO2]
Since the pressure is given, we can use partial pressures in place of concentrations.
At equilibrium, the partial pressure of NO2 is reduced by 0.25 atm (initial pressure) - x, where x is the change in pressure due to the reaction.
The partial pressure of NO will be 2x (according to the stoichiometry).
Using the equilibrium constant expression, we can set up the equation:
K = ([NO]^2 * [O2]) / [NO2]
Substituting the partial pressure values, we get:
5.0 * 10^-13 = (2x)^2 * (P_O2) / (0.25 atm - x)
Now, we need to solve this equation to find the value of x, which represents the change in pressure.
Since x is expected to be very small compared to 0.25 atm, we can assume that (0.25 atm - x) is approximately equal to 0.25 atm. This simplification allows us to solve the equation more easily.
5.0 * 10^-13 ≈ (2x)^2 * P_O2 / (0.25 atm)
Rearranging the equation:
(2x)^2 = 5.0 * 10^-13 * 0.25 atm
4x^2 = 5.0 * 10^-13 * 0.25 atm
x^2 = (5.0 * 10^-13 * 0.25 atm) / 4
x^2 = 1.25 * 10^-13 * 0.25 atm
x^2 = 3.125 * 10^-14 atm^2
Taking the square root of both sides:
x ≈ √(3.125 * 10^-14) atm
x ≈ 5.59017 * 10^-8 atm
Since the pressure of NO2 initially was 0.25 atm, and x represents the change in pressure, the partial pressure of NO2 at equilibrium can be calculated as:
Partial pressure of NO2 at equilibrium = Initial pressure of NO2 - x
= 0.25 atm - 5.59017 * 10^-8 atm
Finally, the partial pressure of NO can be calculated using the stoichiometry, which is twice the value of x:
Partial pressure of NO at equilibrium ≈ 2 * x
≈ 2 * 5.59017 * 10^-8 atm
So, the partial pressure of NO at equilibrium is approximately equal to 1.12 * 10^-7 atm.