5G.3

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Labiba Sardar 2A
Posts: 103
Joined: Sat Jul 20, 2019 12:15 am

5G.3

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
Posts: 119
Joined: Sat Aug 17, 2019 12:15 am

Re: 5G.3

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
Posts: 180
Joined: Wed Sep 18, 2019 12:20 am

Re: 5G.3

partial pressure is used because the reaction only involves gases

Kallista McCarty 1C
Posts: 212
Joined: Wed Sep 18, 2019 12:18 am

Re: 5G.3

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

Jessica Kwek 4F
Posts: 29
Joined: Mon Jan 06, 2020 7:57 am

Re: 5G.3

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
Posts: 109
Joined: Fri Aug 09, 2019 12:15 am

Re: 5G.3

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
Posts: 98
Joined: Wed Sep 18, 2019 12:18 am

Re: 5G.3

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
Posts: 121
Joined: Wed Sep 18, 2019 12:16 am

Re: 5G.3

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
Posts: 79
Joined: Fri Aug 09, 2019 12:16 am

Re: 5G.3

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

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