7th Edition Problem 5.35

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Venya Vaddi 1L
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7th Edition Problem 5.35

Postby Venya Vaddi 1L » Sun Jan 13, 2019 2:42 pm

Does anyone know why all the pressure values are divided by 100 in the K expression in problem 5.35 (7th edition)?

Chem_Mod
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Re: 7th Edition Problem 5.35

Postby Chem_Mod » Mon Jan 14, 2019 1:02 pm

Will you write out the question for those who don't have the 7th edition?

Matthew Tran 1H
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Re: 7th Edition Problem 5.35

Postby Matthew Tran 1H » Mon Jan 14, 2019 7:44 pm

The question showed a graph of the decomposition of compound A into compounds B and C that reach equilibrium. The partial pressure is on the y-axis, in kPa. A changed from a partial pressure of 28 to 18, B from 0 to 5, and C from 0 to 10. Part (a) asked to write the balanced chemical equation for the reaction, but the part in question is part(b), where it asks to calculate K for the reaction. I was initially confused why the pressures had to be divided by 100, but I think it's because the value of Kp depends on the units of pressure. The textbook calculates Kp using atm or bar (1atm ~ 1bar). Since the pressure was given in kPa, you had to convert kPa to atm, and 1kPa ~ 0.01 atm, which is why you had to divide by 100. I'm not sure if we'll be given conversion factors like that on the test though, I had to look it up myself.


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