5J.5

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KNguyen_1I
Posts: 101
Joined: Sat Aug 17, 2019 12:16 am

5J.5

Postby KNguyen_1I » Mon Jan 13, 2020 9:55 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.
(a) 2 O3(g) = 3 O2(g)
(b) H2O(g) + C(s) = H2(g) + CO(g)
(c) 4 NH3(g) + 5 O2(g) = 4 NO(g) + 6 H2O(g)
(d) 2 HD(g) = H2(g) + D2(g)
(e) Cl2(g) = 2 Cl(g)

How do I do this? Isn't it with under the compression of volume the concentrations increase and so the Q quotient then favors going toward the side w the least concentration (eg the one with 'less moles')?

Alicia Lin 2F
Posts: 83
Joined: Wed Sep 18, 2019 12:17 am

Re: 5J.5

Postby Alicia Lin 2F » Mon Jan 13, 2020 10:00 pm

Yes, the composition will want to change in the direction that minimizes the result of increasing pressure and restores equilibrium. There will be less pressure on the side of the reaction with less moles so the composition shifts towards that side.

Isha_Maniyar_Dis2E
Posts: 110
Joined: Thu Jul 11, 2019 12:16 am

Re: 5J.5

Postby Isha_Maniyar_Dis2E » Mon Jan 13, 2020 10:01 pm

a) the reactants are favored
b) the reactants are favored (ignore the carbon because it is a solid, so there are fewer moles of gas on the reactant side)
c) the reactants are favored
d) neither are favored (equal number of moles of gas on both sides)
e) reactants are favored

nicolely2F
Posts: 149
Joined: Sat Sep 14, 2019 12:17 am

Re: 5J.5

Postby nicolely2F » Mon Jan 13, 2020 10:02 pm

The simple way to think about it is the way you mentioned: when pressure is increased by compression, the reaction will tend towards the side with the higher number of moles. This means, for example, that in a) the reactant formation is favored. When the number of moles of gas is equal in both sides, however, there is no change.


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