Reaction Quotient

Reaction Quotient

• Calculated like Keq as reaction proceeds.

Consider the following equilibrium.

A 50.0 L vessel contains 1.00 mol N2, 3.00 mol H2, and 0.500 mol NH3. In which direction (toward reactants or toward products) will the system shift in order to reestablish equilibrium at 400 oC?

The Kc for the reaction at 400 oC is 0.500.

(g )2N H )g(H3 )g(N 322

A Problem to Consider...

First, calculate concentrations from moles of substances.

(g )2N H )g(H3 )g(N 322 1.00 mol

50.0 L3.00 mol

50.0 L0.500 mol

50.0 L


0.0100 M0.0600 M0.0200 M


(g )2N H )g(H3 )g(N 322

The Qc expression for the system would be:

322

23

c ]H][N[

]N H[Q


0.0100 M0.0600 M0.0200 M


(g )2N H )g(H3 )g(N 322

Substituting these concentrations into the reaction quotient gives:

1.23)0600.0)(0200.0(

)0100.0(Q 3

2

c


0.0100 M0.0600 M0.0200 M


(g )2N H )g(H3 )g(N 322

Because Qc = 23.1 is greater than Kc = 0.500, the reaction will go to the left (toward reactants) as it approaches equilibrium.


First, calculate the initial molarities of CO and H2O.

1.000 mol50.0 L

1.000 mol50.0 L

(g )H(g )C O )g(OH)g(C O 222


We must now set up a table of concentrations (starting, change, and equilibrium expressions in x).

First, calculate the initial molarities of CO and H2O.

0.0200 M 0.0200 M

The starting concentrations of the products are 0.

0 M 0 M

(g )H(g )C O )g(OH)g(C O 222


Let x be the moles per liter of product formed.

I 0.0200 0.0200 0 0C -x -x +x +xE 0.0200-x 0.0200-x x x

The equilibrium-constant expression is:

]OH][C O[]H][C O[

K2

22c

(g )H(g )C O )g(OH)g(C O 222


Solving for x.

Substituting the values for equilibrium concentrations, we get:

)x0 2 0 0.0)(x0 2 0 0.0()x)(x(

5 8.0

I 0.0200 0.0200 0 0C -x -x +x +xE 0.0200-x 0.0200-x x x

(g )H(g )C O )g(OH)g(C O 222


Solving for x.

Or:

2

2

)x0200.0(

x58.0

(g )H(g )C O )g(OH)g(C O 222 I 0.0200 0.0200 0 0C -x -x +x +xE 0.0200-x 0.0200-x x x


Solving for x.

Taking the square root of both sides we get:


2

2

)x0 2 0 0.0(

x5 8.0


Solving for x.

Taking the square root of both sides we get:

)x0 2 0 0.0(x

7 6.0

(g )H(g )C O )g(OH)g(C O 222 Initial 0.0200 0.0200 0 0

Change -x -x +x +xEquilibrium 0.0200-x 0.0200-x x x


Solving for x.

Rearranging to solve for x gives:

0 0 8 6.07 6.1

7 6.00 2 0 0.0x



Solving for equilibrium concentrations.

If you substitute for x in the last line of the table you obtain the following equilibrium concentrations.0.0114 M CO0.0114 M H2O 0.0086 M H2

0.0086 M CO2



The preceding example illustrates the three steps in solving for equilibrium concentrations.1. Set up a table of concentrations (starting, change,

and equilibrium expressions in x).

2. Substitute the expressions in x for the equilibrium concentrations into the equilibrium-constant equation.

3. Solve the equilibrium-constant equation for the the values of the equilibrium concentrations.


In some cases it is necessary to solve a quadratic equation to obtain equilibrium concentrations.

Remember, that for the general quadratic equation:

0cb xa x 2 the roots are defined as:

a2a c4bb

x2


In some cases it is necessary to solve a quadratic equation to obtain equilibrium concentrations.

The next example illustrates how to solve such an equation.

Remember, that for the general quadratic equation:

0cb xa x 2


Consider the following equilibrium.

Suppose 1.00 mol H2 and 2.00 mol I2 are placed in a 1.00-L vessel. How many moles per liter of each substance are in the gaseous mixture when it comes to equilibrium at 458 oC?

The Kc at this temperature is 49.7.

H I(g )2 )g(I)g(H 22


The concentrations of substances are as follows.

I 1.00 2.00 0C -x -x +2xE 1.00-x 2.00-x 2x

The equilibrium-constant expression is:

]I][H[]H I[

K22

2

c

H I(g )2 )g(I)g(H 22


The concentrations of substances are as follows.

Substituting our equilibrium concentration expressions gives:

)x0 0.2)(x0 0.1()x2(

7.4 92

I 1.00 2.00 0C -x -x +2xE 1.00-x 2.00-x 2x

H I(g )2 )g(I)g(H 22


Solving for x.

Because the right side of this equation is not a perfect square, you must solve the quadratic equation.

I 1.00 2.00 0C -x -x +2xE 1.00-x 2.00-x 2x

)x0 0.2)(x0 0.1()x2(

7.4 92

H I(g )2 )g(I)g(H 22


Solving for x.

The equation rearranges to give:

00 0.2x0 0.3x9 2 0.0 2

I 1.00 2.00 0C -x -x +2xE 1.00-x 2.00-x 2x

H I(g )2 )g(I)g(H 22


Solving for x.

The two possible solutions to the quadratic equation are:

0 . 9 3x a n d 3 3.2x

I 1.00 2.00 0C -x -x +2xE 1.00-x 2.00-x 2x

H I(g )2 )g(I)g(H 22


Solving for x.

However, x = 2.33 gives a negative value to 1.00-x (the equilibrium concentration of H2), which is not possible.

r e m a i n s 0 . 9 3x o n l y

Initial 1.00 2.00 0Change -x -x +2x

Equilibrium 1.00-x 2.00-x 2x

H I(g )2 )g(I)g(H 22



If you substitute 0.93 for x in the last line of the table you obtain the following equilibrium concentrations.

0.07 M H2 1.07 M I2 1.86 M HI

I 1.00 2.00 0C -x -x +2xE 1.00-x 2.00-x 2x

H I(g )2 )g(I)g(H 22



I 1.00 2.00 0C -x -x +2xE 1.00-x 2.00-x 2x

H I(g )2 )g(I)g(H 22


Reaction Quotient

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