Question

1. 2COF2 yields CO2+CF4 Kc=7.10 If only COF2 is present initially at a concentration of 2M...

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?

Homework Answers

Answer #1


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

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