Gen Chem equilibrium acs QUESTION??

[Philippines03j]

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This question comes from pg 74 number 5. It says

Co + 2H2 <==> Ch3Oh (All gasses)

when .4 mol of c0 and .3 mol of H2 are allowed to reach equilibrium in 1 Liter container, .06 mol of ch3oh are formed. What is the value of Kc?

I know that products go over reactants so it should be .06/(.3^2x.4)=1.7. Also we know at standard situation the liters should be 22.4 and not 1. Because volume is decreased the pressure rises and should shift to the right creating more ch30h which would increase 1.7 to some other higher number.

Answer choices
a. .5 b. .98 c.1.7 d. 5.4

From these answer choices it should be 5.4 but my question is how do you calculate that 5.4. I reasoned through it just dont know how to calculate it. Any help would be appreciated!🙂

 

[nze82]

Since the container is 1L, then the EQ number of moles is also equal to the concentrations.
So:
[CO] = 0.34M
[H2] = 0.18M
[CH3OH] = 0.06M

Keq = [CH3OH]/[H2]^2*[CO] = (0.06)/(0.18^2)(0.34) = 5.4

Hope this helps!

 

[Philippines03j]

Ok, I forgot about the ICE equation. Could you also explain exactly how you calculate the change. I am sure it is easy but its been awhile. Thanks

 

[nze82]

Ok, I forgot about the ICE equation. Could you also explain exactly how you calculate the change. I am sure it is easy but its been awhile. Thanks

Look at the coefficients. The ratio of coefficients of CO:H2:CH3OH is x:2x:x.
We know that when we reach equilibrium, we have 0.06 moles of CH3OH. So:

x = 0.06

Now, we can find the number of moles of CO and H2 consumed in order to produce this many moles of CH3OH, based on the given ratios:

CO = x = 0.06moles
H2 = 2x = 2(0.06) = 0.12moles

Now, if we started with 0.4moles of CO and we have consumed 0.06moles, then at equilibrium we only have 0.4 - 0.06 = 0.34moles.
Same goes for H2: We started with 0.3moles of H2, and we consumed 0.12moles during the process, so at equilibrium we're left with 0.3 - 0.12 = 0.18 moles.

 

[Philippines03j]

Got it! Thanks