Question

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Answer 1

I solved by converting
1324 torr = 1.74 atm
1774 torr = 2.33 atm

I used ICE box.
CO (g) + H2O (g) <-> CO2 (g) + H2 (g)
1. 74 2.33 0 0
-x -x +x +x
1.74-x 2.33-x x x

\[k _{p}=\frac{ [CO _{2}][H _{2}] }{ [CO ][H _{2}O]}\]

\[0.0611 = \frac{ x^2 }{ (1.74-x)(2.33-x) } =\frac{ x^2 }{ x^2-4.07x +4.05 }\]

So, from here...I have no idea what to do next.
Please explain STEP-BY-STEP." A trophy will be given for your guidance.

Note:
If you can do an example problem with an example solution, i can learn it that way. An equation really is not much of a help. I need to see "how" it's done with an example so that I can know how to do it and why it's done step-by-step.

Answer 2

Ignore my first response. My app was not displaying your tex properly

Look at this and tell me if helps

Answer 3

Continuation

Answer 4

it's just algebra.

\(0.0611=\dfrac{x^2}{x^2−4.07x+4.05}\)


\(0.0611(x^2−4.07x+4.05)=x^2\)

\(0.0611x^2−0.248677x+0.247455=x^2\)

\(0.0611x^2-x^2−0.248677x+0.247455=0\)

\(-0.0611x^2−0.248677x+0.247455=0\)

then use the quadratic equation, which gives you

x=-4.8970316210692013, x=0.8270316210692013

i'll let you be the judge of which value for x is right