Textbook question 5J.5

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Paige Lee 1A
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Joined: Sat Sep 07, 2019 12:16 am

Textbook question 5J.5

Postby Paige Lee 1A » Mon Jan 13, 2020 1:39 pm

Could someone please explain why parts D and E have different answers? Why is part E not "no change" also?

Haley Dveirin 1E
Posts: 101
Joined: Sat Jul 20, 2019 12:17 am

Re: Textbook question 5J.5

Postby Haley Dveirin 1E » Mon Jan 13, 2020 2:00 pm

Part E is different from part D (no change) because there is 1 mole of the gas on the reactant side but two moles on the product side (looking at the sum of the coefficients, not the sum of the total number of elements) and therefore the reactants would be favored if the pressure was increased because there are less moles of gas.

Brian Tangsombatvisit 1C
Posts: 119
Joined: Sat Aug 17, 2019 12:15 am

Re: Textbook question 5J.5

Postby Brian Tangsombatvisit 1C » Mon Jan 13, 2020 2:47 pm

Keep in mind that this method of determining the direction of the reaction can only be used if the pressure is increased via compression. If the pressure were to be increased by the addition of inert, noble gases, then this would have no effect of the reaction. The real reason why compression forces the reaction to change is because this compression affects the volume all of the gases (decreasing volume), increasing concentration of each gas since concentration = n/V. Increasing concentration this way would cause the reactant quotient to change, making it differ from K. Depending on whether the new Q > or < K will allow you to explicitly determine which way the reaction will have to proceed to offset the increase in pressure caused by compression.


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