How Do You Calculate the Increase in H2 to Double NH3 at Equilibrium?

[cscott]

A 1.00L vessel contains at equilibrium 0.300 mol of N_2, 0.400 mol H_2 and 0.100 mol NH_3. If the temp is maintained constant, how many moles of H_2 must be introduced into the vessel in order to double the equilibrium concentration of NH_3

I said that if I let the change in conc. of H_2 be equal to x, then the change in conc. of \[NH_3\] would be 2/3x (from the chemical equation). Can't I just equation 2/3x with 0.100? It just seems too easy...

 

[PPonte]

The reaction is:

N_2 + 3H_2 -> 2NH_3

So, in equilibrium we have:
Concentration of N_2: 0.300 mol.dm^-3
Concentration of 3H_2: 0.400 mol.dm^-3
Concentration of 2NH_3: 0.100 mol.dm^-3

With this values you can determine the equilibrium constant K_c. I think you know the formula, if not¹.

But the problem asks how many moles of H₂ must be introduced into the vessel in order to double the equilibrium concentration of NH₃.

Then in the new equilibrium the concentration of each substance must be:
Concentration of N_2: 0.300 + x mol.dm^-3
Concentration of 3H_2: 0.400 + 3x mol.dm^-3
Concentration of 2NH_3: 0.200 mol.dm^-3

Now you apply the equilibrium constant formula and determine x and the solution of the problem.

¹ http://en.wikipedia.org/wiki/Equilibrium_Constant

 

[cscott]

So there will be four solutions?

 

[PPonte]

Four solutions?
How?
The solution to the problem itself should be 3x. And you find x as I explained.
If you don't understand I will resolve the problem entirely.

 

[cscott]

I see now. Thanks for your help.

 

[PPonte]

You're welcome. I hope you check if this is correct with your teacher.

 

[GCT]

PPonte gave some very good points, except the stoichiometries should be modified

Then in the new equilibrium the concentration of each substance must be:
Concentration of : 0.300 + x mol.dm-3
Concentration of : 0.400 + 3x mol.dm-3
Concentration of : 0.200 mol.dm-3

Use factor labeling, and use x for the moles of hydrogen gas introduced.

 

[PPonte]

CGT, you are right. But I just used this stoichiometries in order to simplify the calculations since I prefer to use integral numbers than fractions. If x is the number of moles of hydrogen introduced, the final concentration of N₂ is 0.300 + x/3 M.

 

[MidniteBlaze]

Question: How the heck do you solve for x? I found that K = 0.52

The equilibrium equation would be

K = [2NH_3]^2 / [H_2]^3[N_2]

0.52 = (0.200)^2 / (0.400+3x)^3(0.300+x)

Once you expand that, you get some ridiculous quartic equation. Is there an easier way to solve for x? :S

 

[fishingspree2]

Question: How the heck do you solve for x? I found that K = 0.52

The equilibrium equation would be

K = [2NH_3]^2 / [H_2]^3[N_2]

0.52 = (0.200)^2 / (0.400+3x)^3(0.300+x)

Once you expand that, you get some ridiculous quartic equation. Is there an easier way to solve for x? :S

hey... sorry if the thread is old, but I am having the same problem
I can't solve
0.52 = ((0.2)^2)/((0.3+x)*(0.4+3x)^3)
I get a huge 4th degree polynomial