The reaction : CO (g) + H20 (g) ---> C02 (g) + H2(g) has a Kc value of 4.0 @ 500

Question

The reaction : CO (g) + H20 (g) ---> C02 (g) + H2(g) has a Kc value of 4.0 @ 500 deg C. Calculate the concentration of all species at equilibrium starting with

a) CO & H20 = .100 M
   C02 & H2 = 0.00 M

b) CO, H20, CO2, H2 = .040

* I understand I have to set up the kc equation and do the ICE tables but when I get up to the quadratic equation I get invalid answers! It would be great to see the steps for at least part A just so I can check my math. Thanks!

I set up the ICE table like this

Explanation / Answer

CO (g) + H20 (g) ---> C02 (g) + H2(g)   Kc=4

0.1           0.1              0            0     I

-x              -x               x           x C

0.1-x     0.1-x                x              x    E

Kc = [CO2][H2] / [CO][H2O] = x * x / (0.1-x)^2 = x^2/0.001 - 0.2x + x^2 =4

0.004-0.8x +4x^2 = x^2

3x^2- 0.8x +0.004=0

x= 0.0050974M=[CO2]= [H2]

[H2O]=[CO]=0.1- 0.0050974=0.09490M

CO (g) + H20 (g) ---> C02 (g) + H2(g)   Kc=4

0.04         0.04         0.04          0.04      I

-x             -x              x               x         C

0.04-x     0.04-x       0.04+x      0.04+x       E

Kc = [CO2][H2] / [CO][H2O]

4= [0.04+x]^2/[0.04-x]^2= 1.6x10^-3+0.08x +x^2 / 1.6x10^-3 -0.08 x +x^2

1.6x10^-3+0.08x +x^2 =6.4 x10^-3 - 0.32x +4x^2

3x^2-0.4x +4.8x10^-3 =0

x=.013333

[CO2]=[H2]= 0.04 +.013333= 0.05333M

[CO]=[H2O]= 0.04-.013333=0.02666M

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