1. 2COF2 yields CO2+CF4 Kc=7.10
If only COF2 is present initially at a concentration of 2M what concentration of COF2 remains at equilibrium?
2. CO+NH3 yields HCONH2 Kc=.770
If a reaction vessel initially contains only CO and NH3 at concentrations of 1M and 2M what will be the concentration of HCONH2 at equilibrium?
3. COnsider mixture c which will cause the reaction in reverse
COncentraion xy yields x + y
initial .200 .300 .300
change +x -x -x
equilibrium .200+x .300-x .300-x
based on the Kc value given (.160) and the data table what are the concentration of xy, x, and y at equilibrium?
1.
2COF2 ---> CO2+CF4 Kc=7.10
Kc = [CO2][CF4]/[COF2]^2
7.1 = x^2 / (2-2x)^2
x = 0.842 M
concentration of COF2 at equilibrium = 2-2*0.842 = 0.316 M
2.
CO+NH3 --> HCONH2 Kc=.770
kC = [hconh2]/[co][nh3]
0.77 = x/((1-x)(2-x))
x = 0.53 M
concentration of HCONH2 at equilibrium = 0.53 M
3. HCONH2 ---> CO+NH3 Kc=
0.16
0.16 = (0.3-x)^2 / (0.2+x)
x = 0.086 M
CONCENTRATION OF x at equilibrium = 0.3-0.086 = 0.214 M
CONCENTRATION OF y at equilibrium = 0.3-0.086 = 0.214 M
CONCENTRATION OF xy at equilibrium = 0.2+0.086 = 0.286 M
Get Answers For Free
Most questions answered within 1 hours.