Monday, April 18

We started our day by getting our quizzes back from Friday. Also, due to reasons I forgot, the test on Wednesday will be pushed back to Thursday.

Shortly, we moved on to page 19 of our packet to work on #7. This was different than the other problems we worked on before because this time, it gave us the equilibrium constant (5.76) and the initial concentrations of H2O (8.2) and CO (8.2). The other concentrations were 0 because it was not given. The mole ratios of the equation were all one, so they all had a change in x. The reaction's equilibrium constant was 8.2-x and the product's equilibrium constant was x. To solve this, all you have to do was set up an equation 5.76= x^2 / (8.2-x)^2. To solve for x, you square both sides, getting 2.4= x/(8.2-x). Multiply both sides by 8.2-x and get 19.68 - 2.4x = x. This equation makes x = 5.78. Remember that the problem is asking for the equilibrium concentration, so the answer is not simply x. The answer would be [H2O]=[CO]=2.42 and [H2]=[CO2]=5.78. Next, we moved on to #8 on the same page. This one was a little bit different because it gave us the concentration on both sides. This is quite similar to #7 except that the direction of reaction must be found out. To find the direction, we have to find Q. If you don't remember what Q is, we did some problems on page 16. The Q value was .5166 and was higher than the equilibrium constant of .175, the equation went to the left. After finding that out, everything was basically the same as #7. The answer to #8 is [HCN]=.85 and [C2N2]=[H2]=.355.

As a reminder, these two type of questions will definitely be on the test. Before we moved on to our next topic, we took a little joke break. I don't really remember any of the jokes so...

We moved on to pages 24-25 of our packet. We worked on a and c. In these types of problems, we were given the Ksp and needed to find the equilibrium ion concentration. In problem a we were given that Ksp=1e-10. Because BaSO4 disassociates into Ba and SO4 their initial is 0. They both have a change in x. The Ksp equation is [Ba]x[SO4], so it is x^2=1e-10. This makes x= 1e-5. The answer to problem a is [Ba 2+]=[SO4 2-]=1e-5. Problem c was basically the same except that one of them had a coefficient of 3. Because of this it becomes +3x and in the equation it becomes 27x^3 because the 3 and the x must be cubed. The rest can be solved with the same method as problem a. The answer to problem c is [OH -]= 3x----> 2.0x10^-9=[Al 3+].

We ended our day with us returning the calculators and Mr.H accusing us of trying to steal them.