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# RATE LAW FROM DATA

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CHANG, 7TH EDITION, CHAPTER 14, WORKSHEET #1 S. B. PIEPHO, FALL 2002. HOW TO SOLVE EQUILIBRIUM PROBLEMS. EQUILIBRIUM PROBLEMS ARE A COMMON TYPE OF PROBLEM ...

thChang, 7 Edition, Chapter 14, Worksheet #1 S. B. Piepho, Fall 2002

How to Solve Equilibrium Problems

Equilibrium problems are a common type of problem in Chemistry which involve the equilibrium constant K. More specifically, K is used when the equilibrium constant is written in c

terms of concentrations, and K is appropriate for a gas reaction if partial pressures of gases are p

given.

For a reaction

a A + b B c C + d D

K is defined to be ccd，;[C][D]~;：; Kcab；;?;[A][B](at equilibrium)

where all concentrations are equilibrium concentrations. The double arrow in the equation indicates the reaction goes in both directions (and never reaches completion). At equilibrium, the rates of the forward and reverse reactions are equal, and there is no further change in concentration. K is a constant for a given reaction at a given temperature. c

Methods for solving equilibrium problems are best shown with examples. ______________________________________________________________________________

Example 1 [Similar problems: Chang, #14.76 and 14.82.]

oCalculate the equilibrium constant K at 25 C for the reaction c

2 NOCl(g) 2 NO(g) + Cl(g) 2

using the following information. In one experiment 2.00 mol of NOCl is placed in a 1.00 -L flask, and the concentration of NO after equilibrium is achieved is 0.66 mol/L.

Method:

(a) Calculate the initial concentration of NOCl from the information given. (b) Form an equilibrium table and fill in all known quantities. Represent unknown quantities with variables. Use the information in the table to calculate the concentration of NOCl and Cl at 2

equilibrium.

(c) Form the K equation and fill in all known information. Calculate K. cc

Solution:

Step(a): The initial concentration of NOCl is

2.0 mol [NOCl] = 2.00 M 01.00 L

Step(b): The equilibrium table is:

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thChang, 7 Edition, Chapter 14, Worksheet #1 S. B. Piepho, Fall 2002

Balanced Equation (g) 2 NO(g) + Cl2 NOCl(g) 2

Initial Concentrations (M) 2.00 0 0

Change (M) - 2x 2x x

Equilibrium Concentrations (M) (2.00 - 2x) 2x = 0.66 x

From the table we have

[NO] = 2x = 0.66 M

x = 0.33 M

Thus

[NOCl] = (2.00 - 2x) M = (2.00 - 2(0.33)) M = 1.34 M

[Cl] = x = 0.33 M 2

Step(c): We now have the information we need to calculate K: c

22，;[NO](0.66)[Cl](0.33)~;2：; K0.080c22；;?;[NOCl](1.34)；？at equilibrium

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Example 2 [Related problem: Chang, #14.40.]

When solid ammonium carbamate sublimes, it dissociates completely into ammonia and carbon

dioxide according to the equation

NHCO(s) 2 NH(g) + CO(g) 26232

oAt 25 C, experiment shows that the total pressure of the gases in equilibrium with the solid is

0.116 atm. What is the equilibrium constant K? p

Method:

The steps are:

(a) Form an equilibrium table and fill in all known quantities. Represent unknown quantities with

variables.

(b) Use the table together with the rule that the sum of the partial pressures equals the total

pressure to find the partial pressures of NH and CO at equilibrium. 32

(c) Form the K equation and fill in all known information. Calculate K. pp

Solution:

Step (a): The equilibrium table is:

Balanced Equation (g) + CO(g) 2 NHNHCO(s) 32262

Initial Partial Pressures (atm) (solid!) 0 0

Change (atm) not applicable 2x x

Equilibrium Partial Pressures (atm) not applicable 2x = p(NH) x = p(CO) 32

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thChang, 7 Edition, Chapter 14, Worksheet #1 S. B. Piepho, Fall 2002

Step (b): The total pressure at equilibrium (given as 0.116 atm) equals the sum of the partial pressures. The last line of the equilibrium table allows us to express p(NH) and p(CO) in terms 32

of x. Thus we have

p = p(NH) + p(CO) Total32

0.116 atm = 2x + x = 3x

x = 0.03867 atm

It follows that

p(NH) = 2 x = 2 (0.0387 atm) = .07734 atm 3

p(CO) = x = 0.03867 atm 2

Step (c): Now we have the information needed to calculate K: p

224 Kp p(0.07734)(0.03867)2.31(10；？；？pNHCO32；？at equilibrium

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Example 3 [Similar problems: Chang, #14.42 and 14.46.]

A sample of 0.0020 moles of F was sealed into a 2.0 L reaction vessel and heated to 1000 K to 2

study the dissociation into F atoms:

F 2 F 2

-4(At this temperature, K = 1.210. What are [F] and [F] at equilibrium? What is the percent c2

dissociation of F? 2

Method:

The steps are:

(a) Calculate the initial concentration of F from mole and volume information given. 2

(b) Form an equilibrium table and fill in all known quantities.

(c) Form the K equation and fill in all known information. Solve for the unknown. c

(d) Calculate results asked for in the problem.

Solution:

Step (a): The initial concentration of F is 2

0.0020 mol，;~; [F] = 0.0010 M20；;?;2.0 L

Step (b): The equilibrium table is given below. Note that as the reaction proceeds, reactants are

consumed and products are formed in proportion to their stoichiometric coefficients.

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thChang, 7 Edition, Chapter 14, Worksheet #1 S. B. Piepho, Fall 2002

Balanced Equation 2 F F 2

Initial Concentration (M) 0.0010 0

Change (M) - x 2x

Equilibrium Concentration (M) 0.0010 - x 2x

Step (c): Next we form the K equation and fill in equilibrium values. Note that [F] = (2x) and c

that the entire quantity is squared:

222，;~;[F]2x4x；？4：;?;K1.2(10 c；;[F]?;；？0.0010x；？0.0010x2(at eq)

-42( (0.0010 - x)(1.210) = 4 x

2Next we rearrange the equation above to quadratic form, a x + b x + c = 0:

2-4-7(( 4 x + 1.210 x - 1.210 = 0

In our case,

-4-7(( a = 4, b = 1.210, and c = - 1.210

and, using the quadratic formula, we have:

24427bb4ac(1.2(10)(1.2(10)4(4)(1.2(10)x 2a2(4)

461.2(101.934(10 x8

-4-4((This gives x = 1.5910 and x = - 1.8910. Clearly, only the positive root makes sense +--4(physically; there is no such thing as a (-) concentration! Thus x = x = 1.5910. +

Step (d): Now we are ready to calculate the results asked for in the problem. For this we go

back to the last line of our equilibrium table, and use the value of x calculated above to obtain

-4-4 (( [F] = (0.0010 - x) = (0.0010 - 1.5910) = 8.4102

-4-4(( [F] = 2 x = 2 (1.5910) = 3.210

The equation for percent dissociation is:

quantity dissociated，;~;% dissociation = ( 100%；;?;initial amount

In this case,

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thChang, 7 Edition, Chapter 14, Worksheet #1 S. B. Piepho, Fall 2002

-4，;，;x~;1.59(10~;：;：;?; % dissociation of F = ( 100% ( 100%=2；;?;]0.0010[F；;?;20

= 15.9 %

;

;;； 16 %

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Approximate Methods

In many cases calculations may be simplified by making approximations. For example, if K is c-5(very small (i.e., < 5.010) and the initial reactant concentration c is fairly large (i.e., c > 00

(1000K ), to a good approximation c

(cx)c00

Use of this type of approximation eliminates the need to solve quadratic equations in many cases.

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