[A5HLEY]

Ok, so this is a three part problem, and two of my answers are correct. I can't figure out where I'm going wrong on part C.

Problem:
A flask is charged with 2.780 atm of N2O4(g) and 0.930 atm NO2(g) at 25°C. The equilibrium reaction is given in the equation below:
N2O4(g) <==>  2NO2(g)
After equilibrium is reached, the partial pressure of NO2 is 0.512 atm.

(a) Calculate the equilibrium partial pressure of N2O4: 3.016 atm
(b) Calculate the value of Kp for the reaction: .0869177
(c) Evaluate Kc

Ok, so I thought that Kc = [NO2]^2/[N2O4]
=(.512^2)/(3.016)
=.086917719

but apparently, this is incorrect. Can anyone see what I'm doing incorrectly?

[sjb]

Something as crazy as forgetting units?

[A5HLEY]

Something as crazy as forgetting units?

Haha, no, the program that grades it inserts the units. I only have to provide the number.

[Borek]

Kc = Kp ?

[A5HLEY]

So Kc should just equal .0869177?

[Borek]

No, Kc and Kp are two different things - one uses pressures, other one concentrations.

[A5HLEY]

Ok, so if I need Kc, which deals with concentrations, how do I figure out concentrations of N2O4 and NO2 based on the information given?

[Borek]

Start with concentration definition.

Solve PV=nRT for n. Plug it into above definition.

Calculate Kc using concentrations. You should get something like Kc = something x Kp.

[A5HLEY]

Ok, so I got:

(3.016*1)/*(.0821*298.15) = .1232121019 mol N2O4
(.512*1)/(.0821*298.15) = .0209166433 mol NO2

So then I plugged that into [NO2]^2/[N2O4] = (.0209166344)^2/(.1232121019)
Kc = .003550836 x Kp
= .003550836 x .0869177 = 3.086304981E-4

[A5HLEY]

Does that look correct?

[Borek]

Looks OK to me. But I am known to be occasionally wrong.

[LQ43]

A few clarifications...

1. Please would you check if there is a typo in the initial pressures?

Problem:
A flask is charged with 2.780 atm of N2O4(g) and 0.930 atm NO2(g) at 25°C. The equilibrium reaction is given in the equation below:
N2O4(g) <==>  2NO2(g)
After equilibrium is reached, the partial pressure of NO2 is 0.512 atm.

(a) Calculate the equilibrium partial pressure of N2O4: 3.016 atm

Equilibrium pressure of N2O4 should be 2.989atm but since 3.016 is correct is there a typo in the initial pressures?

2.

Kc = something x Kp.

Kp = Kc(RT)^(delta ngas)               

where delta ngas =  moles gas product - moles gas reactant

3.

Ok, so I got:

(3.016*1)/*(.0821*298.15) = .1232121019 mol N2O4
(.512*1)/(.0821*298.15) = .0209166433 mol NO2

So then I plugged that into [NO2]^2/[N2O4] = (.0209166344)^2/(.1232121019)
Kc = .003550836 x Kp
= .003550836 x .0869177 = 3.086304981E-4

Does that look correct?

Unfortunately not, but just a step too far. You already calculated Kc using the equilibrium moles of N2O4 and NO2

Kc = .003550833 (sig figs?)