Is this right? chem

joejo

Hi guys can you please check if this is right....thanks in advance

Given the reaction CO2(g) + H2(g) = H2O(g) + CO(g). If initially there are 2.0 moles of CO2 (g) and 2.0 moles of H2(g) introduced into an empty 1.0L container, at 1800ºC, it is found that at equilibrium there are 0.3 moles of CO2 still present. Calculate the Ke value for this reaction at 1800ºC.

CO2(g) + H2(g) = H2O(g) + CO(g)
Initially 2 2 0 0
Equilibrium 2-1.7=.3 2-1.7=.3 1.7 1.7

Ke = 1.7 * 1.7/(.3 * .3) = 32.11

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joejo

can anyone help me out plz?

The Bob

I would try but I don't know what Ke is. Sorry.

The Bob (2004 ©)

P.S. Sorry this post was a waste of time.

saltydog

Homework Helper
Alright here goes JoeJo:

This is the equilibrium expression for the reaction:

$$CO_2+H_2\leftrightarrows H_2O+CO$$

$$K_c=\frac{[H_2O][CO]}{[CO_2][H_2]}$$

This at the concentrations at equilibrium which as you indicated, only 1.7 moles of both reagents reacted. So even though it would be nice if you formatted it nicely like I did, your answer looks Ok.

Last edited:

joejo

lol thanks saltydog....i suck at latex...

joejo

shouldn't the answer be 0.03

Vegeta

JoeJo, your answer in #1 is corret, $K_c=32.11$. How do you end up with 0.03?

joejo

Kc=[H2O][CO] over [CO2][H2] = (0.3)^2 over (1.7)^2 = 0.03??

am i right or wrong?

Vegeta

Quote: "it is found that at equilibrium there are 0.3 moles of CO2 still present"
What is $[CO_2]=[H_2]=?$ at equilibrium? By the way you're correct about

$$K_c=\frac{[H_2O][CO]}{[CO_2][H_2]}$$

joejo

$$K_c=\frac{[H_2O][CO]}{[CO_2][H_2]}$$

$$\frac{(0.3)^2}{(1.7)^2}$$

= 0.03

joejo

isn't that right.....my first time with latex!!!

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