Equilibrium constant helpful guide?
Reaction N2 + 3H2<=>2NH3.
If a system of volume 0.3373 mL is prepared from 1.23526 mol of NH3, and at equilibrium there is 0.04171 mol of H2, what would the equilibrium constant be?
I have tried these evil things several times and i never understand the ugly equation. Can someone nicely structure the solution? That way i can also see how you got the answer and then i can do the other sample questions...thanks
Answer:
OK...Are you OK with writing the equation for the equilibrium constant? Since you have the equation written with ammonia on the right and nitrogen and hydrogen on the left, the equilibrium constant equation will be:
K = [HN3]^2 / [N2][H2]^3
In words, the concentration of ammonia squared divided by (the concentration of nitrogen times the cube of the hydrogen concentration)
Since you started with just ammonia and formed 0.04171 moles of hydrogen, you also formed 1/3 that many moles of nitrogen, or 0.01390 moles of nitrogen.
When you formed those compounds, some of the ammonia went away. Using stoichiometry, you can calculate that 0.02781 moles of ammonia disappeared, leaving you with 1.20745 moles of ammonia.
Now, you need to convert each of those moles into a concentration by dividing each one by the volume
[NH3] = 3.5797 mol/mL
[H2] = 0.12366 mol/mL
[N2] = 0.04121 mol/mL
Now, plug those concentrations into the expression for K and solve it.
Kc =
[NH3]^2
--------------------
[N2]^1[H2]^3
Try taking a look at this picture if your having some confusion.
The first thing you have to do is establish the equilibrium compositions. It tells you that at eq. H2 is .04171 mol. It also tells you that you STARTED with 1.23526 mol of NH3. From this you can get eq. compositions of NH3=1.2075, and N2=.0139. Next, use this equation: K=[NH3]^2/([N2]*[H2]^3). This is the concentration of products over concentration of reactants. They are raised to the power of the coefficient in the balanced molecular equation. Plug in the numbers found earlier, and the answer is K = 6.03 x 10^4
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If a system of volume 0.3373 mL is prepared from 1.23526 mol of NH3, and at equilibrium there is 0.04171 mol of H2, what would the equilibrium constant be?
I have tried these evil things several times and i never understand the ugly equation. Can someone nicely structure the solution? That way i can also see how you got the answer and then i can do the other sample questions...thanks
Answer:
OK...Are you OK with writing the equation for the equilibrium constant? Since you have the equation written with ammonia on the right and nitrogen and hydrogen on the left, the equilibrium constant equation will be:
K = [HN3]^2 / [N2][H2]^3
In words, the concentration of ammonia squared divided by (the concentration of nitrogen times the cube of the hydrogen concentration)
Since you started with just ammonia and formed 0.04171 moles of hydrogen, you also formed 1/3 that many moles of nitrogen, or 0.01390 moles of nitrogen.
When you formed those compounds, some of the ammonia went away. Using stoichiometry, you can calculate that 0.02781 moles of ammonia disappeared, leaving you with 1.20745 moles of ammonia.
Now, you need to convert each of those moles into a concentration by dividing each one by the volume
[NH3] = 3.5797 mol/mL
[H2] = 0.12366 mol/mL
[N2] = 0.04121 mol/mL
Now, plug those concentrations into the expression for K and solve it.
Kc =
[NH3]^2
--------------------
[N2]^1[H2]^3
Try taking a look at this picture if your having some confusion.
The first thing you have to do is establish the equilibrium compositions. It tells you that at eq. H2 is .04171 mol. It also tells you that you STARTED with 1.23526 mol of NH3. From this you can get eq. compositions of NH3=1.2075, and N2=.0139. Next, use this equation: K=[NH3]^2/([N2]*[H2]^3). This is the concentration of products over concentration of reactants. They are raised to the power of the coefficient in the balanced molecular equation. Plug in the numbers found earlier, and the answer is K = 6.03 x 10^4
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