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7.3 Equilibrium calculations  (Page 5/11)

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Acetic acid, CH 3 CO 2 H, reacts with ethanol, C 2 H 5 OH, to form water and ethyl acetate, CH 3 CO 2 C 2 H 5 .

CH 3 CO 2 H + C 2 H 5 OH ⇌ CH 3 CO 2 C 2 H 5 + H 2 O

The equilibrium constant for this reaction with dioxane as a solvent is 4.0. What are the equilibrium concentrations when a mixture that is 0.15 M in CH 3 CO 2 H, 0.15 M in C 2 H 5 OH, 0.40 M in CH 3 CO 2 C 2 H 5 , and 0.40 M in H 2 O are mixed in enough dioxane to make 1.0 L of solution?

Answer:

[CH 3 CO 2 H] = 0.36 M , [C 2 H 5 OH] = 0.36 M , [CH 3 CO 2 C 2 H 5 ] = 0.17 M , [H 2 O] = 0.17 M

Check your learning

A 1.00-L flask is filled with 1.00 moles of H 2 and 2.00 moles of I 2 . The value of the equilibrium constant for the reaction of hydrogen and iodine reacting to form hydrogen iodide is 50.5 under the given conditions. What are the equilibrium concentrations of H 2 , I 2 , and HI in moles/L?

H 2 ( g ) + I 2 ( g ) ⇌ 2 HI ( g )

Answer:

[H 2 ] = 0.06 M , [I 2 ] = 1.06 M , [HI] = 1.88 M

Sometimes it is possible to use chemical insight to find solutions to equilibrium problems without actually solving a quadratic (or more complicated) equation. First, however, it is useful to verify that equilibrium can be obtained starting from two extremes: all (or mostly) reactants and all (or mostly) products (similar to what was shown in [link] ).

Consider the ionization of 0.150 M HA, a weak acid.

HA ( a q ) ⇌ H + ( a q ) + A − ( a q ) K c = 6.80 × 10 −4

The most obvious way to determine the equilibrium concentrations would be to start with only reactants. This could be called the “all reactant” starting point. Using x for the amount of acid ionized at equilibrium, this is the ICE table and solution.

This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following: Initial pressure ( M ), Change ( M ), Equilibrium ( M ). The second column has the header, “H A ( a q ) equilibrium arrow H superscript plus sign ( a q ) plus A subscript negative sign ( a q ).” Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.150, negative x, 0.150 minus x. The second column has the following: 0, x, x. The third column has the following: 0, x, x.

Setting up and solving the quadratic equation gives

K c = [ H + ] [ A − ] [ HA ] = ( x ) ( x ) ( 0.150 − x ) = 6.80 × 10 −4
x 2 + 6.80 × 10 −4 x −1.02 × 10 −4 = 0
x = − 6.80 × 10 −4 ± ( 6.80 × 10 −4 ) 2 − ( 4 ) ( 1 ) ( −1.02 × 10 −4 ) ( 2 ) ( 1 )
x = 0.00977 M or −0.0104 M

Using the positive (physical) root, the equilibrium concentrations are

[ HA ] = 0.150 − x = 0.140 M
[ H + ] = [ A − ] = x = 0.00977 M

A less obvious way to solve the problem would be to assume all the HA ionizes first, then the system comes to equilibrium. This could be called the “all product” starting point. Assuming all of the HA ionizes gives

[ HA ] = 0.150 − 0.150 = 0 M
[ H + ] = 0 + 0.150 = 0.150 M
[ A − ] = 0 + 0.150 = 0.150 M

Using these as initial concentrations and “ y ” to represent the concentration of HA at equilibrium, this is the ICE table for this starting point.

This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following: Initial pressure ( M ), Change ( M ), Equilibrium ( M ). The second column has the header, “H A ( a q ) equilibrium arrow H superscript plus sign ( a q ) plus A subscript negative sign ( a q ).” Under the second column is a subgroup of three columns and three rows. The first column has the following: 0, positive y, y. The second column has the following: 0.150, negative y, 0.150 minus y. The third column has the following: 0.150, negative y, and 0.150 minus y.

Setting up and solving the quadratic equation gives

K c = [ H + ] [ A − ] [ HA ] = ( 0.150 − y ) ( 0.150 − y ) ( y ) = 6.80 × 10 −4
6.80 × 10 −4 y = 0.0225 − 0.300 y + y 2

Retain a few extra significant figures to minimize rounding problems.

y 2 − 0.30068 y + 0.022500 = 0
y = 0.30068 ± ( 0.30068 ) 2 − ( 4 ) ( 1 ) ( 0.022500 ) ( 2 ) ( 1 )
y = 0.30068 ± 0.020210 2

Rounding each solution to three significant figures gives

y = 0.160 M or y = 0.140 M

Using the physically significant root (0.140 M) gives the equilibrium concentrations as

[ HA ] = y = 0.140 M
[ H + ] = 0.150 − y = 0.010 M
[ A − ] = 0.150 − y = 0.010 M

Thus, the two approaches give the same results (to three decimal places ), and show that both starting points lead to the same equilibrium conditions. The “all reactant” starting point resulted in a relatively small change ( x ) because the system was close to equilibrium, while the “all product” starting point had a relatively large change ( y ) that was nearly the size of the initial concentrations. It can be said that a system that starts “close” to equilibrium will require only a ”small” change in conditions ( x ) to reach equilibrium.

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Source:  OpenStax, Ut austin - principles of chemistry. OpenStax CNX. Mar 31, 2016 Download for free at http://legacy.cnx.org/content/col11830/1.13
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