### Textbook question 5.35

Posted:

**Mon Jan 13, 2020 2:26 pm**Could someone please explain how to get the correct answers for parts a and b?

Created by Dr. Laurence Lavelle

https://lavelle.chem.ucla.edu/forum/

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=49&t=56199

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Posted: **Mon Jan 13, 2020 2:26 pm**

Could someone please explain how to get the correct answers for parts a and b?

Posted: **Mon Jan 13, 2020 2:39 pm**

a) Set up a quick ice table. We estimate the initial concentrations to be A=27, B=C=0. The change will be: A goes from 28 to 18, B goes from 0 to 5, and C goes from 0 to 10. We can see that the differences between initial and final concentration are in multiples of 5 (A = -10, B = +5, and C = +10). From here, if we pretend that the change, or x, is 5 (in the ice table), then we can say the coefficients for each change are -2, +1, and +2, respectively. (28 - 2x = 18 // 0 + x = 5// 0 + 2x = 10). From ice tables, these coefficients correspond to the stoichiometric coefficients of the chemical equation and we can determine the equation is 2A -> B + 2C. For this problem, you're basically trying to find the ratio of the reactants used / products formed and putting this ratio in the chemical equation.

b) from the correct chemical equation you just found, write the K expression and plug in the equilibrium partial pressures.

b) from the correct chemical equation you just found, write the K expression and plug in the equilibrium partial pressures.

Posted: **Mon Jan 13, 2020 2:44 pm**

For part a you can find the balanced equation by looking at the graph. Substance A decreases from 28 P/kPa to 18 making a difference of 10. Substance B increases from 0 to 5 while Substance C increases from 0 to 10. This means that 10 P/kPa of A was lost -> 10 P/kPa of C and 5 P/kPa was created. If we put this into stoichiometric values we get 2A <--> B + 2C. For Part B, we take these values and plug them into the equilibrium constant. To get the same Kp value as the answer key, you'll need to convert from kPa to atm.