Homework 11.79 6th edition

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Julia Lung 1I
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Homework 11.79 6th edition

Postby Julia Lung 1I » Sat Jan 12, 2019 1:52 pm

What is the significance of using bar for units and what would be the easiest approach for problems like 11.79 6th edition?

11.79 A reactor for the production of ammonia by the Haber process is found to be at equilibrium with PN2  3.11 bar, PH2  1.64 bar, and PNH3  23.72 bar. If the partial pressure of N2 is increased by 1.57 bar, what will be the partial pressure of each gas once equilibrium is re-established?

Aarti K Jain 1L
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Re: Homework 11.79 6th edition

Postby Aarti K Jain 1L » Sat Jan 12, 2019 2:23 pm

There really is no significance for using bar for units as opposed to pascals or atm. The textbook writers probably just want to make sure that the reader can use all the different kinds of units and is consistent when using each one.

For this particular problem, I would suggest finding K(p) using the equilibrium partial pressures given to you by the problem. Then, I would add the 1.57 bar to the partial pressure of N2. With the initial values becoming P(N2) = 4.68 bar, P(H2) = 1.64 bar, and P(NH3) = 23.72 bar, you can just create an ICE table and use the K(p) value to find what the new equilibrium values will be.

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Re: Homework 11.79 6th edition

Postby Sean_Rodriguez_1J » Sat Jan 12, 2019 2:37 pm

In regards to the first part, the use of bar as your units is not really significant; the bar is just a certain kind of unit of pressure (note that bar is a non-SI unit) and it won't affect how we solve the problem.

To start the problem, calculate K based on the equilibrium pressures that are given so that would look like
K = (23.72)^2/((3.11)(1.64)^3) = 41.0. The exponents are there to account for the stoichiometric coefficients needed to balance the equation.
With an established K, you can then make an ICE chart.
N2 + 3H2 -> 2NH3 All of these are gases at the temperature of the reaction, so we won't need to change the bar units. Be sure to remember that the stoichiometric coefficients affect the changes in concentration.
I 4.68 bar 1.64 bar 23.72 bar
C -x -3x +2x
E 4.68-x 1.64 - 3x 23.72 + 2x
Now we can set the K we found earlier equal to the expression made from our ICE chart because K will remain the same.
41.0 = (23.72 + 2x)^2/((4.68-x)(1.64-3x)^3)
This seems pretty hard to do by hand, so a solver on your calculator or a graphing calculator will get you x=0.0656 as an answer that makes sense. To calculate the pressures...
PN2 = 4.68 - 0.0656 = 4.61 bar, PH2 = 1.64 - 3(0.0656) = 1.44 bar, and PNH3 = 23.72 + 2(0.0656) = 23.85 bar.
Hope this helps :)

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