Homework question 5J.5

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sbeall_1C
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Homework question 5J.5

Postby sbeall_1C » Fri Jan 17, 2020 1:05 pm

For question 5J.5, many reactions are given and we are asked to predict whether the reactants or products would be favored with an increase in total pressure. One of the reactions is 2HD(g)+H2(g)⇌D2(g). The answer says no change, but wouldn't the reaction favor the products because there are less moles of gas? Another reaction is Cl2(g)⇌2Cl(g) and the answer says the reaction would favor the reactants, but wouldn't this one be no change?

Thanks!!

Bella Townsend
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Re: Homework question 5J.5

Postby Bella Townsend » Fri Jan 17, 2020 1:46 pm

for the reaction 2HD(g)+H2(g)⇌D2(g), H2 is an inert gas and it does not affect the reaction. Typically noble gases in reactions are inert gases.


the reaction Cl2(g)⇌2Cl(g) goes from 1 mole Cl in the reactants, and 2 moles Cl in the products. Le chatelier's principles says that the reaction goes to the side with fewer molecules of gas, so the reaction moves toward the reactants.

Chris Tai 1B
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Re: Homework question 5J.5

Postby Chris Tai 1B » Fri Jan 17, 2020 4:16 pm

In part d of this question, the equation produced is actually 2HD(g)⇌H2(g)+D2(g) instead of 2HD(g)+H2(g)⇌D2(g), which makes sense since the equation is now balanced. I'm not entirely sure how all the inert gas/noble gas stuff affects the way that the reactants/products shift in the event of higher pressure, but this makes the answer of "no change" make sense to me since the amount of moles of gas are equivalent on both sides of the equation.

Pegah Nasseri 1K
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Re: Homework question 5J.5

Postby Pegah Nasseri 1K » Fri Jan 17, 2020 5:41 pm

When pressure increases, the volume of the container decreases. The molar concentrations of the reactants and products change due to this change in volume. It is helpful to remember that if pressure increases, the reaction shifts to the side with fewer moles. For part (d) of this question, there are 2 moles of HD on the reactant side, and one mole of H2 and one mole of D2 on the product side. The amount of moles on the reactant and product side are equal, so the reaction won't shift at all and will not change. Meanwhile, in part (e) there is one mole of Cl2 on the reactant side and 2 moles of Cl on the product side so the reaction will shift to the reactant side because it has fewer moles.


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