HW 5J.5 part C

Moderators: Chem_Mod, Chem_Admin

Posts: 106
Joined: Fri Aug 30, 2019 12:16 am

HW 5J.5 part C

Postby CNourian2H » Wed Jan 15, 2020 6:26 pm

5J.5 State whether reactants or products will be favored by an increase in the total pressure (resulting from compression) on each of the following equilibria. If there is no change, explain why that is so.

(c) 4NH3(g)+ 5O2(g) -> HNO(g) + 6H2O(g)

How do I know the reactants are favored in this case? I am confused on the step to step work that would solve this.

Hannah Lee 2F
Posts: 117
Joined: Thu Jul 11, 2019 12:15 am

Re: HW 5J.5 part C

Postby Hannah Lee 2F » Wed Jan 15, 2020 6:50 pm

The equation for (c) is as follows: 4NH3(g)+ 5O2(g) -> 4NO (g) + 6H2O(g)
There are 9 moles of gas on the reactant side and 10 moles of gas on the reactant side.
Compression implies that pressure (and thus concentration) increases while volume decreases. To counteract this increase in pressure/concentration, the rxn would shift in the direction that produces less moles of gas, which in this case is the reactant side.

You can also work this out quantitatively by plugging in sample numbers to prove that Q > K, which causes the reaction to shift left, similar to what we did in class.

Ipsita Srinivas 1K
Posts: 50
Joined: Mon Jun 17, 2019 7:24 am

Re: HW 5J.5 part C

Postby Ipsita Srinivas 1K » Wed Jan 15, 2020 6:50 pm

1.Assume volume of all reactants and products are 0.1 M. (you can take any value!)
Then, Kc = [H2O]6[NO]4/[O2]5[NH3]4
Kc= (0.1)6(0.1)4/(0.1)5(0.1)4
= 0.1.
2. As pressure is increased by compression i.e volume decreases, we can say that the local concentrations of all the reactants and products has increased. This is because PV = nRT. This implies, P = (n/V)(RT). As R,T, and n are constant, and increase in P will result in a decrease in V (they are inversely proportional!). As n/V is the molar conc., if for a fixed n, the denominator V decreases, (n/V) value will increase.

3. Say V decreased by 1/2. Then, the new (n/V) values will be twice their original values aka 2x(0.1) = 0.2 mol/L.

4. Now calculate Q using the new conc.
Qc = (0.2)10/(0.2)9 = 0.2 (Just substitute 0.2 for 0.1 in the Kc equation)

5. Q = 0.2; K = 0.1. This means Q>K. As both Q and K are basically [P]/[R], to reduce the Q value, you need a larger denominator value i.e a larger reactant concentration. This means, to get back to equilibrium (the K value!) we need to produce more reactants (cause we're right now at Q) hence the backwards reaction is favoured.

Alexis Webb 2B
Posts: 124
Joined: Thu Jul 11, 2019 12:15 am

Re: HW 5J.5 part C

Postby Alexis Webb 2B » Wed Jan 15, 2020 6:52 pm

The question states for part c: 4NH3 (g) + 5O2 (g) <-> 4NO (g) + 6H2O (g) Since there are more moles of gas on the right, the reaction will shift to the left towards the reactants when there is compression. I think you just forgot some of the information in the problem.

Return to “Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions”

Who is online

Users browsing this forum: No registered users and 1 guest