Help on Chemistry: Equilibrium Constants and Composition?

Question

The equilibrium constant for the reaction N2(g) + 1/2O2(g) <=> N2O(g) is 7.1x10^-19 at 25 degrees(C). The equilibrium constant for the reaction N2(g) + O2(g) <=> 2NO(g) is 4.32x10^-31 at 25 degrees(C).

a. Given this data, what is the equilibrium constant for the reaction below at 25 degrees(C)?

N20(g) + 1/2O2(g) <=> 2NO(g)

b. What is the value of Kc at 25 degrees(C)?

c. Assume the system is in equilibrium in a 1.000L container, if the volume of the container is now reduced to .500L, what happens to the composition of the container?

Any help on this would be simply fantastic!

Answer 1

a.

A. N2 (g) + 1/2O2 (g) &lt;=&gt; N2O (g)..........k1 = 7.1x10^-19

B. N2(g) + O2(g) &lt;=&gt; 2NO(g) ................k2 = 4.32x10^-31

C. N20(g) + 1/2O2(g) &lt;=&gt; 2NO(g)...........k3 = ?

You have to manipulate the first two reactions so that when you add them together they will equal the overall reaction (C).

Since N2O is one of the reactants in reaction C and reaction A is the only reaction that contains N2O we need to reverse reaction A. We don&#039;t have to change anything about reaction B.

N2O (g) &lt;=&gt; N2 (g) + 1/2O2 (g)........k = (1/7.1x10^-19)

(Since we reversed the reaction we had to invert the equilibrium constant)

So now you add reactions A and B to get C:

A. N2O (g) &lt;=&gt; N2 (g) + 1/2O2 (g)

B. N2(g) + O2(g) &lt;=&gt; 2NO(g)

--------------------------------------...

C. N20(g) + 1/2O2(g) &lt;=&gt; 2NO(g).......k3 = (1/k1)(k2) = 6.085 x 10^-13

b.  

You need to use the relationship between kp and kc to find kc

kp = kc(RT)^delta(n)

where n is the sum of the stoichiometric coefficients of the gaseous products minus the sum of the stoichiometric coefficients of  gaseous reactants.

I don&#039;t know what kp is. I think its a constant that you should have been given. Find Kp and plug it in and solve. R is 8.314 j(mol*k) and T is 298K. n is 2 - 1.5 = 0.5 (I think!)

c.

In this question you apply what you know about Le Chatelier&#039;s Principle. When you decrease the volume of the reaction container you shift the equilibrium in the direction that has the fewer moles of gas. so for the reaction N20(g) + 1/2O2(g) &lt;=&gt; 2NO(g if you were to decrease the volume of the reaction container you would shift the equilibrium to the left (reactants) because there are 1.5 moles of gas on the left and 2 moles of gas on the right.

Answer 2

For the given rxn, we have:

  [NO]^2 / [O2]^0.5  [N2O] = Kc

  We can include the K rxns we know

[NO]^2 = 4.32x10-31 [N2][O2]   and

[N2O] = 7.1x10-19 [N2]sqrt[O2]

Inserting these in, everything cancels out except the constants, so Kc= 4.31x10-31/7.1x10-19 =6x10-13 appx.