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### Topic: Units to use for the calculation of Kp and Qp  (Read 5626 times)

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#### confusedstud

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• Mole Snacks: +3/-0 ##### Units to use for the calculation of Kp and Qp
« on: July 12, 2015, 06:57:22 AM »
When we are calculating Qp or Kp for the equation ΔG=ΔG°+RTlnQp what should the units of Qp be?

Say for this reaction CH3OH(g)  <---> CO(g)+2H2(g) given PCH3OH=0.855atm, PCO=0.125atm and PH2=0.183atm. So I'm able to calculate the Qp based on atm giving me 4.90x10^-3 atm^2. However I'm not sure which units to use when substituting it into the equation. If I were to convert the 4.90x10^-3atm^2 to Pa I would get 49x10^6Pa^2.

But when substituting into the lnQp term which value should I use? Because the lnQp term does not have a unit. However substituting either values would yield different results so I'm not very sure which value to substitute into.

#### mjc123

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• Mole Snacks: +229/-11 ##### Re: Units to use for the calculation of Kp and Qp
« Reply #1 on: July 13, 2015, 05:31:36 AM »
Strictly speaking, the value of Q you use in the equation should be dimensionless, as you can only take the log of a number, not a physical quantity. For this purpose you express the pressures as dimensionless numbers, by dividing the pressure by the standard pressure P°. If P° = 1 atm, then PMeOH/P° = 0.855, and so on, and Qp = 4.90 x 10-3. If you're working in Pa, you divide by the value of P° in Pa, so you should get the same number. Similarly for Qc, you divide the concentrations by the standard concentration (1M). "Standard" means whatever is defined as standard in the definition of ΔG°.

#### confusedstud

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• Mole Snacks: +3/-0 ##### Re: Units to use for the calculation of Kp and Qp
« Reply #2 on: July 13, 2015, 07:07:36 AM »
Strictly speaking, the value of Q you use in the equation should be dimensionless, as you can only take the log of a number, not a physical quantity. For this purpose you express the pressures as dimensionless numbers, by dividing the pressure by the standard pressure P°. If P° = 1 atm, then PMeOH/P° = 0.855, and so on, and Qp = 4.90 x 10-3. If you're working in Pa, you divide by the value of P° in Pa, so you should get the same number. Similarly for Qc, you divide the concentrations by the standard concentration (1M). "Standard" means whatever is defined as standard in the definition of ΔG°.

Ohh no wonder in the problem they used atm to solve for since P°=1atm. So for that question where i had 49x106Pa2 So I would have to divide that by (105)(105)2/105 which also gives me 4.9x10-3.

Thanks so much for clearing this up. But actually during my lessons in school when we learned about the equation Kc=Kp(RT)^Δn we had units for both Kc and Kp where the units were atm and M. Is there a reason for this?

#### mjc123

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• Mole Snacks: +229/-11 ##### Re: Units to use for the calculation of Kp and Qp
« Reply #3 on: July 13, 2015, 08:33:01 AM »
Historically, in developing the theory of equilibria, it was found that as a matter of fact (for example) PCO*PH22/PCH3OH = constant (the law of mass action). The value of the constant depends on the units used.
Practically, it is useful to continue using equilibrium constants with units, as it serves as a check that you've got the stoichiometry right, and are using the right units (e.g. not using pressures in Pa with a K value in atm). Reading through these forums shows that people frequently make such mistakes.