Equilibrium Assignment Answers PDF

Preview

Summary

Equilibrium spreadsheet assignment ANSWERS spring 2019, assignment 2 from Peter Warburton...

Description

Chem 1051 Spreadsheet Assignment on Chemical Equilibrium Due Date: Friday July 5th 2019 at 9:00 am Instructions: Write your name and MUN# on your completed assignment, which would include the workings for all questions, and the complete printed spreadsheets and graphs for the last two problems. Each of the equilibrium problems should be completed using the following steps: a)

b) c) d)

e)

Organize the problem by creating an ICE (Initial-Change-Equilibrium) table. Use the letter x to signify any unknown quantity in the problem. If the initial reaction mixture has both reactants and products in it, be sure to do a reaction quotient calculation to correctly identify the direction of reaction. Based on your ICE table, identify the “most negative” possible value x can take and justify how you got that value. Based on your ICE table, identify the “most positive” possible value x can take and justify how you got that value. Determine the equilibrium mixture composition using the method specified in the problem. If you feel comfortable making assumptions about the size of x relative to the initial concentrations to simplify the math, then do so, but you must state your assumption and then check whether the assumption is reasonable at the end (5% rule). If it is not, then the problem must be redone without simplifying assumptions. Check if you have done the problem correctly by doing a reaction quotient calculation using your equilibrium mixture values. If the value of Q does not equal K (within reason), you have most likely made an error somewhere.

PLEASE NOTE: The first two questions can be solved like the example ICE table problem done in class with a quadratic equation. The last two problems will require you to set up a spreadsheet correctly for graphing and solving purposes. 1)

Myoglobin (a protein symbolized by Mb) reacts with oxygen to form a myoglobinoxygen complex Mb (aq) + O2 (g) ⇌ MbO2 (aq)

Kc = 8.6 x 105 at 38 C

The initial reaction mixture contains 2.0 x 10-4 M of Mb and 1.9 x 10-5 M of O2. Find the equilibrium mixture composition for this problem by using appropriate mathematics. [10%] 2)

The equilibrium constant in terms of concentration at 25 C for the reaction CO2 (g) + H2 (g) ⇌ H2O (l) + CO (g) is 3.22 x 10-4. The initial reaction mixture contains 0.68 M of CO2 and 0.95 M of H2. Find the equilibrium mixture composition for this problem using appropriate mathematics. [10%]

3)

The equilibrium constant in terms of concentration at 25 C for the reaction VO+ (aq) + 2 H+ (aq) ⇌ H2O (l) + V3+ (aq) is 14.0. The initial reaction mixture contains 0.090 M of V3+, 0.100 M of H+, and 0.050 M of VO+. i) Solve this problem by using the following method of successive approximations. This part will require you to find the fully expanded polynomial equation of the form ax3 + bx2 + cx + d = 0 which you should then rearrange to have x on one side of the equation and everything else on the other side of the equation. With rearrangement the equation

should look like x = (ax3 + bx2 + d) / -c Estimate a starting value of x that falls between the most negative and most positive values. Show a sample calculation to calculate a “new” value for x by substituting

1

this estimate for x into the right-hand side of x = (ax3 + bx2 + d) / -c and getting the “new x” value on the left-hand side. Set up a spreadsheet with two labelled columns Old x and New x. The first value of Old x should be your estimate from your sample calculation. Set up the calculation to find New x for this first attempt. It should match what you calculated in your sample calculation. Next, tell the spreadsheet that the next Old x value is the previously found New x value, and use this for the second New x calculation by dragging down the previous cell from the column. Finally, highlight the second row of “x values” and drag them down so you have 25 successive calculations (The cells from A1 to B26 should be occupied, with the first row having the labels). Format the values of x in both columns to show four decimal places, and then comment on the value x converged to in the successive approximations. Use this value of x to determine your equilibrium mixture and do a reaction quotient calculation to confirm that it is indeed an equilibrium mixture. [20%] ii) Using the non-expanded form of your equilibrium constant expression (the equilibrium constant expression equation that comes directly from the ICE table before you do any math) use a spreadsheet to do reaction quotient Q calculations (formatted to two decimal places) for values of x between the “most negative” and “most positive” possible values of x [parts b) and c)] at intervals of 0.01. Show the sample calculation for the “most negative” value of x. Graph (XY scatter plot with data points only) the values of Q that are between 0 and 30 versus the value of x and add a sixth order polynomial trendline through the points. Print the graph and draw a line across the graph for the value Q = K, and where it crosses the polynomial curve, draw a line straight down to intersect the x axis. Comment on the value of x and how it compares compare that value to the value you obtained in part i). [20%] iii) Using the fully expanded polynomial from part i) expressed in terms of ax3 + bx2 + cx + d = y calculate values of y (to three significant figures) for values of x between the smallest and largest possible values of x [parts b) and c)] at intervals of 0.01. Again, you only need to show the first calculation, and can then tabulate the rest. Graph the values of y versus the values of x and use a sixth order polynomial trendline through the points. Using this curve estimate what value of x will give a y value of zero (when y is zero, then we must be at equilibrium since this equation would match what we solved for in part i)) and compare that value to the value you obtained in part i). [20%] 4)

The equilibrium constant in terms of concentration at 25 C for the reaction 6 ClO3F (g) ⇌ 2 ClF (g) + 4 ClO (g) + 7 O2 (g) + 2 F2 (g) is 28.0. The initial reaction mixture contains 1.00 M of ClO3F. Create the ICE table, determine the smallest and largest possible values of x, and estimate the value of x while explaining your logic. Then, using the non-expanded form of your equilibrium constant expression (like part ii) of Question 3) calculate values of Q (to three significant figures in scientific notation) for values of x between the smallest and largest possible values of x at intervals of 0.0025 until you get to the first point where Q is larger than K. Show the sample calculation for the first point after the “most negative” value of x. Graph (XY scatter plot with smooth line and markers) the values of Q versus the value of x (a trendline won’t work here). Print the graph showing the range of calculated Q values from 0 to 30. Now, change the bounds of your graph to show the curve between Q values of 27 and 29 and suitable bounds for the values of x so the line occupies most of the graph. Print the graph and draw a line across the graph for the value Q = K, and where it crosses the polynomial curve, draw a line straight down to intersect the x axis to estimate the value of x required to establish equilibrium to five decimal places. Perform a reaction quotient calculation

with this value of x and comment on how well this procedure described the equilibrium. [20%]

2

1. a) A suitable ICE table for this question would look like (all in molL-1) Initial conc. Conc. change Equil. conc.

Mb (aq) + -4 2.0 x 10 -x 2.0 x 10-4 – x

MbO2 (aq) 0.0 +x +x

⇌ O2 (g) -5 1.9 x 10 -x 1.9 x 10-5-x

b) Since we start completely with reactants and no products, the smallest value x can take is 0.00 mol/L. In other words, if the initial mixture happens to be an equilibrium mixture, then no change is needed for the system to get to equilibrium. c) Since we start completely with reactants and no products, the largest value x can take is 1.9 x 10-5 mol/L. If the system is not at equilibrium, there will be some change. The largest amount of change will occur if the reaction proceeds to completion from left to right. This would mean that whatever reactant we run out of first (the limiting reagent) determines the maximum amount of change. In this case the limiting reagent is O2, and so the largest value of x will be the same as the initial concentration of O2. d) To solve for x we substitute our equilibrium concentrations into the equilibrium constant expression K = 8.6 x 10

5

=

[MbO

]

2

[Mb][O

2

so

]

8.6 x 10

5

(x)

= (2.0 x 10

-4

− x)(1.9 x 10

-5

− x)

Rearranging

(8.6 x 10 ( 8.6 x 10 8.6 x 10  8.6 x 10

5 5 5 5

)(2.0 x 10 − x)(1.9 x 10 − x) = x )(3.8 x 10 − 2.19 x 10 x + x ) = x  x - 1.88 x 10 x + 3.27 x 10 = x  x - 1.89 x 10 x + 3.27 x 10 = 0 -4

-5

-9

-4

2

2

2

-3

2

2

-3

This is a quadratic equation of the form ax2 + bx + c = 0, which has solutions given by x =

−b 

so

2a 1.89 x 10 +

2

5

-3

(− 1.89 x 10 ) − 4(8.6 x 10 )(3.27 x 10 ) 5

3.57 x 10 + 1.12 x 10

2

1.89 x 10 + 1.56 x 10 1.72 x 10

x=

2

− (− 1.89 x 10 ) 

4

1.72 x 10 x =

x =

2(8.6 x 10 ) 2

x =

2

2

b − 4ac

3.45 x 10

2

1.72 x 10

6

6

or x =

4

2

6

4

1.89 x 10 −

or x =

3.57 x 10 + 1.12 x 10 1.72 x 10

2

2

1.89 x 10 − 1.56 x 10

or x =

1.72 x 10 0.33 x 10

2

1.72 x 10

6

so

4

6

2

6

x = 2. 00 x 10

−4

-1

mol  L

or

x = 1 .92 x 10

−5

mol  L

-1

Now we have a problem here because both answers will give us a negative concentration for O2 at equilibrium. Another way to see this is that both possible answers are larger than the largest value of x we established in part c). Either we’ve made a math error, or rounding errors make it difficult to solve this problem using the quadratic formula. If your calculator does quadratics, you might not have run into this problem, but if you did, then be sure not to round any of your values until the absolute end. We really needed to solve the for the quadratic formula

( 8.6 x 10 ) (2.0 x 10 − x)(1.9 x 10 − x) = x ( 8.6 x 10 ) (3.8 x 10 − 2.19 x 10 x + x ) = x 8.6 x 10  x - 1.8834 x 10 x + 3.268 x 10  = x 5

-4

5

-9

-5

-4

2

5

2

2

-3

5

2

2

-3

 8 6 x 10  x - 1 8934

x 10

x + 3 268

x 10

= 0

This is a quadratic equation of the form ax2 + bx + c = 0, which has solutions given by

3

x =

−b 

x =

so

2a 1.8934 x 10 +

4

1.72 x 10 1.8934 x 10 + 1. 5687 x 10 1.72 x 10 x=

2

5

-3

(− 1.8934 x 10 ) − 4(8.6 x 10 )(3.268 x 10 ) 5

3.58496 x 10 + 1.12419 x 10

2

x =

2

− (− 1.8934 x 10 )

2(8.6 x 10 ) 2

x =

2

2

b − 4ac

3.4621 x 10 1.72 x 10

6

6

4

2

or x =

6

or x =

3.58496 x 10 + 1.12419 x 10 1.72 x 10

2

2

or x =

1.8934 x 10 − 1.5687 x 10 1.72 x 10

2

4

1.8934 x 10 −

0.3247 x 10 1.72 x 10

4

6

2

6

2

so

6

x = 2. 013 x 10

−4

-1

mol  L

or

x = 1. 888 x 10

−5

mol  L

-1

By avoiding the rounding errors, we do have a value of x that falls within the range of expected change, namely x = 1.888 x 10-5 M. Using these values, the equilibrium concentrations would be [Mb] = 2.0 x 10-4 M - 1.888 x 10-5 M = 1.8112 x 10-4 M [O2] = 1.9 x 10-5 M - 1.888 x 10-5 M = 1.2 x 10-7 M [MbO2] = 1.888 x 10-5 M Using a Q calculation to confirm this is an equilibrium mixture gives Q =

[MbO 2 ] [Mb][O

] 2

=

(1.888 x 10 -5 ) -4

-7

= 8. 7 x 10 5

(1.81 12 x 10 )(1.2 x 10 )

Since Q = K, within rounding errors, we have a reasonable equilibrium mixture. There is an alternative way to answer this question. Let’s use what we know to our advantage. Since we know x will be very close to the largest value possible (since K is very large and we expect a near complete reaction), let’s pretend the reaction first does go all the way to completion. The new reaction mixture after this complete reaction occurs is then allowed to come to equilibrium. So, to put together a new ICE table requires two steps. First, we need to imagine what the reaction mixture looks like if our initial mixture reacts to completion. This is a limiting reagent problem where we figure out which reactant we run out of first! Mb (aq) + O2 (g) ⇌ MbO2 (aq) (all in molL-1) 2.0 x 10-4 1.9 x 10-5 Initial conc. 0.0 Change to get to completion -1.9 x 10-5 -1.9 x 10-5 +1.9 x 10-5 Final conc. 2.0 x 10-4 – 1.9 x 10-5 1.9 x 10-5-1.9 x 10-5 0.0+1.9 x 10-5 =0 = 1.81 x 10-4 =1.9 x 10-5 Now we’ll let this new mixture go to equilibrium by setting up an ICE table where the new initial mixture is the final mixture of the limiting reagent problem. (all in molL-1) Initial conc. Conc. change Equil. conc.

Mb (aq) + -4 1.81 x 10 +y 1.81 x 10-4 + y

O2 (g) 0 +y +y

⇌

MbO2 (aq) 1.9 x 10-5 -y 1.9 x 10-5-y

I’m using y in this case so we understand the change we’re looking for is not the same as x from the original ICE table. If we carefully look at the approach we’re taking here, what we’re saying is that x from the original problem is the largest value we thought it could be (the reaction goes to completion) minus some small amount of change y (we go slightly back towards reactants to reach equilibrium). Overall, we mean x = 1.9 x 10-5 M – y We’ll solve for this new equilibrium problem the exact same way as any other equilibrium

problem

4

K = 8.6 x 10 5 =

[MbO 2 ] [Mb][O

2

(1.9 x 10

8.6 x 10 5 =

so

]

(1.81 x 10

-5

-4

− y)

+ y)(y)

Let’s assume that y is much less than 1.9 x 10-5 M (and therefore is also much less than 1.81 x 10-4 M as well), which seems a reasonable assumption since we expect the equilibrium mixture to have relatively small amounts of the reactants because K is very large. Another way of stating this assumption is that we believe 1.9 x 10

-5

M - y 1.9 x 10

-5

M

and

1.81 x 10

-4

M + y 1.81 x 10

-4

M

This simplifies our math to (1.9 x 10 -5 )

8.6 x 10 5 =

-4

(1.81 x 10 )(y) (1.9 x 10 - 5 )

y=

-4

5

= 1.22 x 10 − 7 M

(1.81 x 10 )(8.6 x 10 )

We need to check our assumption! Using the 5% rule, we see that y 1.9 x 10

−5

= M

1.22 x 10

−7

1.9 x 10

−5

M

= 0.0064 = 0.64%

M

Our assumption was justified. Based on this new ICE table, our equilibrium mixture is [Mb] = 1.81 x 10-4 M + 1.22 x 10-7 M = 1.81122 x 10-4 M [O2] = 1.22 x 10-7 M [MbO2] = 1.9 x 10-5 M - 1.22 x 10-7 M = 1.8878 x 10-5 M Using a Q calculation to confirm this is an equilibrium mixture gives Q =

[MbO 2 ] [Mb][O

] 2

=

(1.8 878 x 10 -5 ) -4

-7

= 8. 5 x 10 5

(1.81122 x 10 )(1.22 x 10 )

Since Q = K, within rounding errors, we have a reasonable equilibrium mixture. Also notice how this approach gives a very similar equilibrium mixture to the quadratic equation approach.

5

2. a) A suitable ICE table for this question would look like CO2 (g) 0.68 -x 0.68 – x

(all in molL-1) Initial conc. Conc. change Equil. conc.

+

H2 (g) 0.95 -x 0.95 - x

H2O (l) N/A N/A N/A

⇌

+

CO (g) 0.0 +x +x

b) Since we start completely with reactants and no products, the smallest value x can take is 0.00 mol/L. In other words, if the initial mixture happens to be an equilibrium mixture, then no change is needed for the system to get to equilibrium. c) Since we start completely with reactants and no products, the largest value x can take is 0.68 mol/L. If the system is not at equilibrium, there will be some change. The largest amount of change will occur if the reaction proceeds to completion from left to right. This would mean that whatever reactant we run out of first (the limiting reagent) determines the maximum amount of change. In this case the limiting reagent is CO2, and so the largest value of x will be the same as the initial concentration of CO2. d) To solve for x we substitute our equilibrium concentrations into the equilibrium constant expression K = 3.22 x 10

−4

[CO]

=

so

[CO 2 ][H 2 ]

3.22 x 10

−4

(x)

=

(0.68 − x)(0.95 − x)

Rearranging

( 3.22 x 10 ( 3.22 x 10  3.22 x 10  3.22 x 10

-4 -4

-4

) (0.68 − x)(0.95 − x) = x )(0.646 − 1. 63 x + x ) = x  x - 5.2486 x 10  x +  2.080 x 10  = x  x -1.00052486  x + 2.080 x 10 = 0 2

2

-4

-4

-4

2

-4

Notice that we have done absolutely no rounding of any of the numbers to avoid rounding errors like we saw in the first approach of Question 1. This is a quadratic equation of the form ax2 + bx + c = 0, which has solutions given by x =

x =

− b

2

b − 4ac

so

2a 1.00052486 +

1.00104999 5+ 2. 67904 x 10

1.00052486 + 1.00052472 6 6.44 x 10

x =

2.00104958 6 6.44 x 10

2

-4

-4

-4

−7

or x =

or x =

9.8 x 10

-4

-4

)

1.00052486 −

1.00104999 5+ 2. 67904 x 10 6.44 x 10

−7

-4

1.00052486 − 1.00052476 2 6.44 x 10

or x =

-4

(− 1.00052486 ) − 4(3.22 x 10 )(2.080 x 10 ) 2(3.22 x 10

6.44 x 10 x =

− (− 1.00052486 )

x=

-4