Chemical Equilibrium Part 2 Question 24

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Adams Dis 1A
Posts: 23
Joined: Wed Nov 14, 2018 12:23 am

Chemical Equilibrium Part 2 Question 24

Postby Adams Dis 1A » Wed Jan 09, 2019 9:19 pm

On this particular question, I am extremely confused about the 65 bars measurement, and how to begin the calculation for Q, any help would be appreciated.

Jonas Talandis
Posts: 60
Joined: Fri Sep 28, 2018 12:16 am

Re: Chemical Equilibrium Part 2 Question 24

Postby Jonas Talandis » Wed Jan 09, 2019 9:22 pm

Bars are units of partial pressure. Treat them the same way you would concentration in the product/reactant expression for Kc, only this time, it's Kp.

Jessica Chen 1F
Posts: 61
Joined: Fri Sep 28, 2018 12:17 am

Re: Chemical Equilibrium Part 2 Question 24

Postby Jessica Chen 1F » Wed Jan 09, 2019 10:33 pm

You can use the equation Q = [(pC)^c(pD)^d] / [(pA)^a(pB)^b] and plug in the partial pressures into A, B, C, D and the stoichiometric coefficients into a, b, c, d.


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