Equillibrium constants?

Question

Okay another question

A mixture of .200 mol NO2 and .200 mol CO is placesd in a 1.00 L flask and given time to equilibriate. Analysis of the equillibrium mixture indicates that 0.134mol of CO2 is present calculate Kc for the reaction.

NO2+CO ---> NO + CO2

the answer is 4.1

I am unsure how to figure it out. please help.

Answer 1

first figure the concentration of each
[NO2]=.2/1=.2M and [CO]= .2/1= .2 M is initial concen.
[CO2]= .134/1= .134M that is concentration when it gets equilibrium
from chemical rxn you can wirte the Kc expression
Kc=[NO][CO2]/([NO2][CO])


write the i-chart
NO2 + CO -> NO + CO2

Initial .2 ..... .2 ......... 0 .... 0 ( cuz no product)
change -x .......-x ......... +x ... +x
Equi .2-x.......2-x .........x ..... x
but they said when it gets equilibrium we see the .134 of CO2 present
you see on the last row of chart is Equil that is equilibrium which means [CO2]=x = .134M( cuz they given that is presented when it gets equilibrium)
so { NO2}= .2-x= .2-.134 cuz x=.134, and {CO}=.2-.134
{NO}= x = .134, { CO2}= x= .134
Kc at equilibrium = (.134*.134)/((.2-.134) *(.2- .134) = 4.122
good luck

Answer 2

Write out the balanced equation:

NO2 + CO ---> NO + CO2

Then write a chart below it with what you know. Since the Kc is used for molarity (moles/L), change your values to molarity. In this case it's easy because the volume of the flask is just 1L, so all the values stay the same. Remember that according to your balanced equation, you use up 1 mole each of NO2 and CO to make 1 mole each of NO and CO2.

Initial concentrations: NO2 = CO = 0.200 M, NO = 0 M, CO2 = 0M
Change in concentration: NO2 = CO = -x, NO = +x, CO2 = +x
Equilibrium concentration: NO2 = CO = 0.2 -x, NO = x, CO2 = 0.134 M

Since the final concentrations of NO and CO2 must be equal (the equation shows that you make the same amount of CO2 and NO from the starting reactants), the equilibrium concentration of NO must also be 0.134 M.

That means that the equilibrium concentrations of NO2 and CO must be 0.2 - .134 = 0.066 M

For the chemical equation aA + bB --> cC + dD, the Kc equation is

Kc = ([C]^c[D]^d)/([A]^a[B]^b)

with the equilibrium concentrations of the products divided by the equilibrium concentrations of the reactants. You have to use the coefficients as powers of the concentrations, but this problem is easy because all the coefficients in the balanced equation are 1.

So for your problem, Kc = ([NO]*[CO2])/([NO2]*[CO]) = (0.134*0.134)/(0.066*0.066) = 4.1

Answer 3

Whatever is formed on the right must be lost from the left (all the ratios are 1:1).
So (0.2 - 0.134) moles of NO2 and CO are left at equilibrium, and 0.134 mol of NO is also formed. The vol is 1 litre, so the concentrations are numerically equal to the moles.

Now write the Kc expression and put the equilibrium concentrations in.

Answer 4

K(eq) = [NO][CO2]/(NO2][CO])
[CO2] = 0.200 - 0.134 = 0.066
[NO2] = 0.200 - 0.134 = 0.066
K(eq) = 0.134 x 0.134 / (0.066 x 0.066) = 4.12
CO can only come from CO2 (1 mole = 1 mole) and the starting conc. of CO2 was 0.200.
If 0.134 mole of CO2 reacts to form 0.134 mole of CO, then 0.134 mole of NO2 had to react to form 0.134 mole of NO