# Am I doing these Chemical Equilibrium Mass Action Expression problems correctly?

1. Apr 17, 2005

### DLxX

I need help with the following questions. If you don't have time to solve both then please do the 2nd one. The answers I got are

1a .5
1b 1
1c not sure how to do it

2a 3.86
2b 2.42^13

Are they correct?

1) For the following reaction

2SO2 + O2 <_-_-_-_-_-> 2SO3

What is the value of the Mass Action Expression when 2.5 moles of SO2 and 1.0 moles of O2 are placed in a 500mL container? What is the value after .50 moles of the O2 have been used up? What is the value at equilibrium if there is still 1.5 moles of O2 left?

4NH3 + 5O2 <_-_-_-_-_-> 4NO + 6H2O

What is the value of the mass action expression in (d) above if, in a 2.5 liter container there is 7.5 moles of NH3, 5.0 moles of O2, 25.0 moles of NO and 2.5 moles of H2O? What is the value of the mass action expression if, at equilibrium, the concentration of the NO has increased to 11M?

Last edited: Apr 17, 2005
2. Apr 17, 2005

### chem_tr

My help will not be about solving your problems, you don't seem to show your brainstorming here.

Subscripts are done in two ways in PF. First way is using BBCode technique. In here, use {sub} and {/sub} to indicate a subscripts, like this: H{sub}2{/sub}O. The only error I made intentionally is to use waved paranthesis instead of the correct one, capped paranthesis []. Please change them with [] to obtain H2O.

Second way is to use LaTeX technique, which is similar and sometimes much more good-looking and easier. It needs you to memorize some codes, though. In LaTeX technique, subscripts are made with _ sign, if you need to show more than one characters as subscript, then enclose them between {}. Start and end the LaTeX code with {tex} and {/tex}, respectively; here, don't forget to use the correct paranthesis, []. The result is like this: $$H_2O$$

Superscripts are done with a similar way, in BBCode technique, use {sup} and {/sup}, and in LaTeX technique, use the character ^ with {}, if needed.

I want to say that you will have to provide your query a bit more understandable to attract some attention and increase your chance to get a response. Sorry if I am being rude and offensive.

Good luck.

Last edited: Apr 17, 2005
3. Apr 17, 2005

### DLxX

Thank you. I've changed my original post.

4. Apr 18, 2005

### chem_tr

Your first reaction includes the following conditions, if I understood correctly (the volume is constant and 0.5 L):

a) 2.5 moles of SO2 vs. 1.0 mole of O2
b) 0.5 moles of O2 is used
c) 1.5 moles of O2 is excess in the medium (equilibrium is established)

If the reaction is like below, let's try to find them:

$$2SO_2 + O_2 \rightleftharpoons 2SO_3$$

In a), 0.5 moles of SO2 will be excess.

In b), both SO2 and O2 will be excess at 1.5 and 0.5 moles, resp., with occurrence of 1.0 mole of SO3.

In c), if 1.5 moles of O2 is excess, what is the new value of SO2? I think this part of the question is unsolvable unless we know additional parameters.

You know the mass action expression, $$\frac{concentration~of~products}{concentration~of~reactants}$$. The volume is 0.5 L, so just divide the moles by 0.5 to learn their concentrations, and find a and b. I am again not sure about c.

Well, this one looks a bit tougher than the former, but is still solvable.

You will first try to find which one has been used up completely, and rely on this to find the reacted amounts of the others, to find, at last, how many moles of them are present when the equilibrium has been reached.

First, if there are 7.5 moles of ammonia and 25.0 moles of NO, and the coefficients of them in the reaction are both 4, so the equilibrium will probably shift to left. In addition, water seems to be the least in the medium, so it will probably be used up completely. So let's rely on this to find the others.

If 2.50 moles of water is completely used up, then 1.67 moles of NO will also be used, so 23.33 moles of NO will be present. Since the coefficients are same, 1.67 moles of NH3 will be formed, and since there are already 7.50 moles of ammonia, the final mass will be 9.17 moles. In addition, about 2.08 moles of oxygen will be formed, and the total of oxygen gas will be 7.08 moles. By using these values, you'll calculate the mass action expression, I think. A brief warning: I may do some errors, but if you understand the process, you can repeat it with no errors and calculate the real value.

Edit: Of course, the amount of water will not be zero; so let's say the remaining amount of water is 2.50-x, so the others will be 23.33-x (nitrogen monoxide), 9.17+x (ammonia) and 7.08+x (oxygen). The equilibrium constant will help you to find x.

Last edited: Apr 18, 2005