5G.3

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Labiba Sardar 2A
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5G.3

Postby Labiba Sardar 2A » Mon Jan 06, 2020 6:06 pm

Why is K written in terms of the partial pressures of the reactants and products rather than the concentrations of the reactants and products for both parts 5G.3 a) and b)?

Brian Tangsombatvisit 1C
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Re: 5G.3

Postby Brian Tangsombatvisit 1C » Mon Jan 06, 2020 7:07 pm

In this case, since all the reactants and products for both reactions are in the gas phase, it is easier to derive the Kp if given the units for the gases. This is because Kp is another way to write the equilibrium constant, but for a reaction with all gases since each reactant and product will have a partial pressure that you would plug into Kp.

You do have the option to write the formula for Kc, but if you were given partial pressures to plug in, you would first have to convert all the gases' partial pressures to concentrations using (n/V) = P/RT to be able to use your expression for Kc. Kp is generally used for a reaction only involving gases.

805097738
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Re: 5G.3

Postby 805097738 » Mon Jan 06, 2020 7:24 pm

partial pressure is used because the reaction only involves gases

Kallista McCarty 1C
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Re: 5G.3

Postby Kallista McCarty 1C » Mon Jan 06, 2020 7:28 pm

This is because the reaction only involves gasses so partial pressures are used!

Jessica Kwek 4F
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Re: 5G.3

Postby Jessica Kwek 4F » Mon Jan 06, 2020 7:36 pm

K is written in terms of partial pressures since the reactants and products for both reactions all in gas phase. This means that you would write the expression for the equilibrium constant K as a formula for Kp. This includes the partial pressure of the product in the numerator, raised to a power equal to its stoichiometric coefficient in the balanced equation. The reactants and products should be interpreted as PJ (numerical value of the pressure in bar).

anjali41
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Re: 5G.3

Postby anjali41 » Mon Jan 06, 2020 8:23 pm

Since both parts a and b consist of only molecules in the gas phase, partial pressures are used throughout instead of concentrations. So, when solving for the K value, you would be solving for the Kp value more specifically.

DesireBrown1J
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Re: 5G.3

Postby DesireBrown1J » Mon Jan 06, 2020 11:29 pm

I wrote my answer for part a as K=[C2H4Cl2]2[H2O]2/[C2H4]2[O2][HCl]4 but the textbook has it in terms of partial pressures because they are gases. This is also how we had to write it when doing the modules. If I do not write it in terms of partial pressures on the exams or homework if it doesn't say to write it that way, will I be deducted points?

MAC 4G
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Re: 5G.3

Postby MAC 4G » Mon Jan 06, 2020 11:53 pm

DesireBrown1J wrote:I wrote my answer for part a as K=[C2H4Cl2]2[H2O]2/[C2H4]2[O2][HCl]4 but the textbook has it in terms of partial pressures because they are gases. This is also how we had to write it when doing the modules. If I do not write it in terms of partial pressures on the exams or homework if it doesn't say to write it that way, will I be deducted points?


I don't think you'll lose points for it on the homework because most TAs grade based on completion, but this is what happened last quarter, so I'm not 100% it's the same this quarter. But, to be safe, I'd do it in terms of partial pressure being it's how the answer are written in the book as well as the modules. Generally, since this is a gas-phase reaction, you want the K value in terms of partial pressure.

stephaniekim2K
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Re: 5G.3

Postby stephaniekim2K » Tue Jan 07, 2020 11:18 am

Partial pressure is used in this case because they are gases.


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