# Is this right? chem

1. Jun 18, 2005

### 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

2. Jun 18, 2005

### joejo

can anyone help me out plz?

3. Jun 18, 2005

### The Bob

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

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

4. Jun 18, 2005

### saltydog

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: Jun 18, 2005
5. Jun 18, 2005

### joejo

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

6. Jun 22, 2005

### joejo

7. Jun 22, 2005

### Vegeta

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

8. Jun 22, 2005

### joejo

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

am i right or wrong?

9. Jun 23, 2005

### 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

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

10. Jun 23, 2005

### joejo

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

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

= 0.03

11. Jun 23, 2005

### joejo

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