the total pressure of n2o4 and no2 IS 1.38 atm. if kp is 6.75(25 C) calculate partial pressure of NO2 in the mixture. 2NO2<---> n2o4
I assume the system is at equilibrium?? If so, the way I see it is
total pressure = 1.38 atm..
partial pressure N2O4 = x
partial pressure NO2 = 2x
3x = 1.38 and solve for x, then
2x = partial pressure NO2.
I don't think that is the right interpretation for the problem.
2NO2 ==> N2O4
If total P is 1.38, then
pNO2 = x
pN2O4 = 1.38-x
Then Kp = 6.75 = pN2O4/p(NO2)^2
Substitute x and 1.38-x into the above Kp expression and solve for x.
Nothing
Good
Hi I am retro
To calculate the partial pressure of NO2 in the mixture, we can use the expression for the equilibrium constant (Kp) and the given total pressure of the gases.
The equation for the equilibrium reaction is:
2NO2 ⇌ N2O4
The equilibrium constant (Kp) for this reaction is given as 6.75 (at 25°C).
Kp = (Partial Pressure of N2O4) / (Partial Pressure of NO2)^2
We are given the total pressure of N2O4 and NO2 in the mixture as 1.38 atm. Let's consider the partial pressure of NO2 as "x" atm.
The partial pressure of N2O4 can be calculated as the difference between the total pressure and the partial pressure of NO2:
Partial Pressure of N2O4 = Total Pressure - Partial Pressure of NO2
Partial Pressure of N2O4 = 1.38 atm - x atm
Now, we can substitute these values into the expression for Kp:
Kp = (1.38 atm - x atm) / (x atm)^2
Since we are given a specific value for Kp (6.75), we can rearrange the equation and solve for x:
6.75 = (1.38 - x) / x^2
Rearranging the equation gives:
6.75x^2 = 1.38 - x
6.75x^2 + x - 1.38 = 0
Now, we can solve this quadratic equation to find the value of x using factoring, completing the square, or the quadratic formula. I'll use the quadratic formula to solve for x:
x = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = 6.75, b = 1, and c = -1.38.
x = (-(1) ± √((1)^2 - 4(6.75)(-1.38))) / (2(6.75))
By solving this equation, we can find the values of x. Substitute the values for a, b, and c into the equation, calculate the discriminant, and then solve for x.