## 5J.5

Connor Ho 1B
Posts: 102
Joined: Sat Aug 17, 2019 12:17 am

### 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.

d) 2 HD(g) + H2(g) ⇌ D2(g)

Why is the answer no change? Is it because the H2 omitted on the product side? Why is it only omitted on one side of the equation?

Ryan Lee 1E
Posts: 50
Joined: Sat Aug 17, 2019 12:16 am

### Re: 5J.5

I agree with you that it seems the 2 H2 molecules were just omitted for some reason.

805422680
Posts: 103
Joined: Sat Sep 14, 2019 12:16 am

### Re: 5J.5

the H2 molecule has been omitted from the right side of the equation. if it were present, the number of molecules would be the same on the right and the left of the equation, therefore, there'd be no change in equilibrium as a result of a change in pressure.

Posts: 111
Joined: Thu Jul 25, 2019 12:17 am

### Re: 5J.5

I can answer why there is no change from the increase in pressure. Usually, in a Kp equation, the partial pressure is representative of the number of mole/concentrations of a substance at a certain temperature and volume. You can think of it as how much "stuff" is there increases with pressure. For example, if you add a compound in a flask of the same volume and at the same temperature, the partial pressure of that specific compound you added will increase (because you have more of that particle). That extra compound you added, will then affect your equilibrium equation (since you added more "stuff" either reactant or product).

In this case, it says that the pressure increases due to compression. So the amount of "stuff" is the same, the flask or container just got smaller. Therefore, the reaction will stay at equilibrium or do whatever it was going to do before.

I'm not quite sure what's up with the missing Hydrogens...

Jamie Lee 1F
Posts: 106
Joined: Fri Aug 09, 2019 12:16 am

### Re: 5J.5

The problem should state 2HD ⇌ H2 + D2, I think you misplaced a reactant on the product side :)

So with this, there would be no change since there is an equal amount of moles of gas on either side.