What is the Ke value for the given reaction at 1800ºC?

[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

 

[joejo]

can anyone help me out please?

 

[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]

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.

 

[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]

= (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!