Equilibrium Partial Pressure Kp/Kc Question

In summary: P=nRT/V, what have you used for n?Thanks for the help, I did indeed misplace a decimal xD.n=.0821/.0552=3.14
  • #1
YeyFunNOT
4
0

Homework Statement


At 100 o C Kc=.078 for the reaction SO2Cl2<-->SO2 + Cl2. In an equilibrium mixture the [SO2CL2]=.0108 M and [SO2]=.052 M.

What is the partial pressure of Cl2 in the eq. mixture?

Homework Equations


Kp=Kc(RT)[itex]\Delta[/itex]n
P=RT/V

The Attempt at a Solution


I solve for Kp but I don't know how to find the partial pressure since neither the total pressure nor the volume is given.
 
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  • #2
Pressure can't be a function of the volume, which suggests in the end it will cancel out. Would it help to assume 1L total volume?
 
  • #3
I tried that, here's the work:
P=RT=(.0821)(373)=30.6233 atm

Moles is M*L so all of the moles equal the concentration. Total number of moles is .0108+.052+.0162 (moles of Cl2)=.079 total moles. So the pCl2= .0162/.079*30.6233=6.27 atm.

The book says 5 atm so where am I going wrong?
 
  • #4
YeyFunNOT said:
I tried that, here's the work:
P=RT=(.0821)(373)=30.6233 atm

List assumptions used to calculate this number.

But I don't see how the final answer can be 5 atm. Are you sure you have not missed a decimal point?
 
  • #5
Borek said:
List assumptions used to calculate this number.

But I don't see how the final answer can be 5 atm. Are you sure you have not missed a decimal point?
I used the ideal gas law, PV=nRT rearranged to solve for P. (I did this in another problem in the same set to get overall pressure). I just set the volume as 1 L.

I checked all the numbers and the answer in the book and everything seems correct. :S
 
  • #6
YeyFunNOT said:
I used the ideal gas law, PV=nRT rearranged to solve for P. (I did this in another problem in the same set to get overall pressure). I just set the volume as 1 L.

P=nRT/V, what have you used for n?
 
  • #7
Thanks for the help, I did indeed misplace a decimal xD.
 

FAQ: Equilibrium Partial Pressure Kp/Kc Question

What is equilibrium partial pressure Kp/Kc?

Equilibrium partial pressure Kp/Kc is a measure of the partial pressure of a gas at equilibrium in a chemical reaction. It is a ratio of the products' partial pressures to the reactants' partial pressures, and is used to determine the direction and extent of a reaction.

How is equilibrium partial pressure Kp/Kc calculated?

To calculate equilibrium partial pressure Kp/Kc, you must first write the equilibrium expression for the reaction. Then, plug in the partial pressures of the products and reactants at equilibrium and solve for Kp/Kc. Alternatively, you can use the ideal gas law to calculate the partial pressures if the temperature and volume are known.

What is the difference between Kp and Kc?

Kp and Kc are both equilibrium constants, but they are calculated using different units. Kp is calculated using partial pressures, while Kc is calculated using molar concentrations. This means that Kp is affected by changes in volume, while Kc is not.

How does temperature affect equilibrium partial pressure Kp/Kc?

According to Le Chatelier's principle, an increase in temperature will shift the equilibrium of an endothermic reaction to the right, increasing the partial pressures of the products and thus, increasing Kp/Kc. On the other hand, a decrease in temperature will shift the equilibrium of an exothermic reaction to the left, decreasing Kp/Kc.

What are some real-life applications of equilibrium partial pressure Kp/Kc?

Equilibrium partial pressure Kp/Kc is used in various industries, such as in the production of ammonia for fertilizers and in the Haber process for the production of nitric acid. It is also used in the production of industrial chemicals, such as methanol and ethanol. Additionally, Kp/Kc is used in environmental chemistry to understand the behavior of pollutants and in pharmaceuticals to determine the optimal conditions for drug synthesis.

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