total pressure equilibrium chemistry help!?
at -80oC, Kc for the reaction N204 <==> 2NO2 is 4.66e-8. We introduce 0.013 mole of N2O4 into a 1.0L-vessel at -80oC and let equilibrium be established. The total pressure in the system at equilibrium will be?
i dont know how to start this problem at all. I tried the ICE table but got really confused.
Plz help explain this problem. Thank you so much!!
Comments
Kp = Kc(RT)^delta n
This is the equation you must use. Kp = gas pressure
K = 273 + -80 = 193K
delta n = moles product - moles reactant; 2-1 = 1
Kp=(4.66x10^-8Moles/L(0.0821L-atm/mol-K)(193K)e1
Kp = 7.38x10^-7atm
properly, you recognize which you're commencing with .1mol SO2, .1mol O2, and no SO3. whilst the reaction is done, you will have (.a million mol - 2x) SO2, (.1mol - x) O2, and 2x SO3. Plug this into your Ksp and resolve for x. The Ksp could be [SO3]^2/[SO2]^2[O2]=.355 So, you will have: (2x)^2/(.a million-2x)^2(.a million-x)=.355 No you are able to resolve for x. i decide for to propose simplifying so as which you have an equation that =0 (so it is going to likely be ax^4+bx^3+....-.355=0). Graph it with a graphing calculator and notice the place it hits the x-axis. which would be your answer for x.