5.33

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vpena_1I
Posts: 81
Joined: Sat Aug 24, 2019 12:15 am

5.33

Postby vpena_1I » Wed Jan 08, 2020 7:55 pm

Dissociation of a diatomic molecule, X2(g)<-->2 X(g) occurs at 500 K. The equilibrium state of the reaction is shown in 1 and the equilibrium state in the same container after a change has occurred is shown in 2. Which of the following changes will produce the composition shown? (a) Increasing the tempera- ture. (b) Adding X atoms. (c) Decreasing the volume. (d) Adding a catalyst. Explain your selections.


(container 1 shows 8 X2 molecules and 6 X molecules. container 2 shows 2 X2 molecules and 18 X molecules)

The answer is a, with the reasoning being that the reaction is endothermic. Why is the reaction endothermic? I thought breaking bonds released energy.

VPatankar_2L
Posts: 89
Joined: Thu Jul 25, 2019 12:17 am
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Re: 5.33

Postby VPatankar_2L » Wed Jan 08, 2020 8:49 pm

Hi! Energy is actually required to break bonds which means that a certain amount of energy must be provided to the reaction in order for the dissociation of X2 to occur.
As you can see in this image (https://i.stack.imgur.com/jm5uO.png) bound atoms have low potential energy, and energy is required for the bond between them to break. External energy was needed for the atoms shown at the lowest point on the graph to separate, as shown by the two split atoms at the far right.

Sara Richmond 2K
Posts: 99
Joined: Fri Aug 30, 2019 12:16 am

Re: 5.33

Postby Sara Richmond 2K » Sat Jan 11, 2020 3:26 pm

In the case of this problem, why doesn't reducing the volume result in the production of more X molecules? The solution manual says that it would result in the production of more X2 molecules which doesn't make sense to me as in the equation there are more moles on the right side of the equation.

805303639
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Re: 5.33

Postby 805303639 » Wed Jan 15, 2020 9:15 pm

Hi Sara. Reducing the volume increases the concentrations of both X2 and X (concentration = n/V). According to the reaction quotient Q = [X]^2/[X2], products increase relative to reactants as a result of the concentration increases. Q > K and the system favors the reverse reaction, yielding more X2 molecules.

In short, I think you had the logic for the short-cut approach reversed: compression favors the side with fewer moles of gas.


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