Given the initial concentrations of 0.10 atm NO2 and 0.10 atm N2O4 in a 1.0 L flask, what will be the equilibrium partial pressure of NO2?
N2O4(g)=2NO2 Kp=0.660at319K
Choose one answer.
a. 0.10 atm
b. 0.31 atm
c. 0.045 atm
d. 0.72 atm
e. 0.19 atm
0.19
Surely you jest. Concentrations are not measured in atm.
To find the equilibrium partial pressure of NO2, we first need to set up the equilibrium expression for the given reaction.
The balanced equation for the reaction is:
N2O4(g) ⇌ 2NO2(g)
The equilibrium constant expression for this reaction is:
Kp = [NO2]^2 / [N2O4]
We are given the initial partial pressures of NO2 and N2O4 as 0.10 atm each. Note that the equilibrium constant (Kp) value is also given as 0.660.
Now we can use the given information and the equilibrium constant expression to calculate the equilibrium partial pressure of NO2.
Let's assume that at equilibrium the partial pressure of NO2 is x atm. Since the stoichiometric coefficient of NO2 in the balanced equation is 2, the partial pressure of N2O4 at equilibrium will be (0.10 - x) atm.
Substituting these values into the equilibrium constant expression, we get:
0.660 = (x^2) / (0.10 - x)
Next, we need to solve this equation for x to find the equilibrium partial pressure of NO2.
To do this, we rearrange the equation:
0.660(0.10 - x) = x^2
Expanding and rearranging the equation further, we get:
0.066 - 0.66x = x^2
Rearranging and setting the equation equal to zero:
x^2 + 0.66x - 0.066 = 0
Now we can use the quadratic formula to solve for x:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a=1, b=0.66, and c=-0.066.
Solving this equation, we get two possible values for x:
x1 = 0.0463 atm (approximately)
x2 = -0.7063 atm (approximately)
Since we are dealing with partial pressures, the concentration of NO2 cannot be negative. Therefore, we discard the negative value.
So, the equilibrium partial pressure of NO2 is approximately 0.0463 atm.
Comparing this value to the answer choices provided, we can see that the closest option is:
c. 0.045 atm