### 11.117

Posted:

**Sun Dec 03, 2017 9:19 pm**Does anyone know how to solve exercise 11.117 for the equilibrium chapter?

Created by Dr. Laurence Lavelle

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

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=49&t=24421

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Posted: **Sun Dec 03, 2017 9:19 pm**

Does anyone know how to solve exercise 11.117 for the equilibrium chapter?

Posted: **Sun Dec 03, 2017 9:37 pm**

Hello!

For part a)

i) The amount of NO2 will increase because if you increase the amount of NO the reaction will proceed toward the products in order to minimize the effects of the change

ii) Removing the SO2 will also cause the reaction to proceed toward the products, therefore producing more NO2

iii) As discussed in class, if you increase the pressure by adding gas the mols of the reactants and products remain constant and there is no change in their concentration meaning there is no effect on the amount of NO2

For part b)

I did it a different way then the book says because it did not make sense to me, but I ended up with the correct answer.

you are given that there are .240 moles of SO2 at equilibruim. NO2 and SO2 will have equal values because their ratio is 1:1.

SO3 + NO -> SO2 + NO2

I. .245 y. 0. 0

c. -x. -x. +x. +x

e. (.245-x)(y-x). (x). (x)

therefore, x= .240. SO3 would be equal to .005 and NO would be (y-.240)

You would set up the equation to solve for y...

K=6.0x10^3= (.240)(.240)/ (.005)(y-.240)

For part a)

i) The amount of NO2 will increase because if you increase the amount of NO the reaction will proceed toward the products in order to minimize the effects of the change

ii) Removing the SO2 will also cause the reaction to proceed toward the products, therefore producing more NO2

iii) As discussed in class, if you increase the pressure by adding gas the mols of the reactants and products remain constant and there is no change in their concentration meaning there is no effect on the amount of NO2

For part b)

I did it a different way then the book says because it did not make sense to me, but I ended up with the correct answer.

you are given that there are .240 moles of SO2 at equilibruim. NO2 and SO2 will have equal values because their ratio is 1:1.

SO3 + NO -> SO2 + NO2

I. .245 y. 0. 0

c. -x. -x. +x. +x

e. (.245-x)(y-x). (x). (x)

therefore, x= .240. SO3 would be equal to .005 and NO would be (y-.240)

You would set up the equation to solve for y...

K=6.0x10^3= (.240)(.240)/ (.005)(y-.240)

Posted: **Sun Dec 03, 2017 9:38 pm**

sorry the alignment for the ICE box got messed up! Hopefully you can still understand it :-)