The Meaning of K_c

When we were doing stoichiometric calculations early in the course, we assumed that the reactions ran 100% in the forward direction. Another way to say this is that we assumed the reactions "went to completion." Many reactions reach an equilibrium point long before they reach the 100% mark. The size of the equilibrium constant gives a measure of how the reaction proceeds.

In order to interpret the meaning of K_c, it will be helpful for us to think about the equilibrium expression (written above). Here we have written it in an abbreviated form.

$LaTeX: K_c=\frac{\text{[products]}}{\text{[reactants]}}$

(This abbreviation is useful for thinking about K_c conceptually, but remember when you actually calculate K_c that you will need to include exponents from the balanced equation.)

Look at the abbreviation equation for K_c to answer the following.

When the concentration of the products is larger than the concentration of reactants what value of K_c will you get?
- Answer: (larger than 1)
When the concentration of the products is smaller than the concentration of reactants, what value of K_c will you get?
- Answer: (less than 1)
When the concentration of the products equals the concentration of reactants, what value of K_c will you get?
- Answer: (equals 1)

equilibrium diagram of K>>1, K=1, and K<<1 with mix of products and reactants in each

The size of K_c also gives us information about the position of the equilibrium, or how far toward completion the reaction goes before a state of equilibrium is established.

K_c is very small

Extremely small amount of products formed

Position of equilibrium favors reactants

K_c≈1

Position of equilibrium lays midway toward completion

K_c is very large

Very large amount of products formed

Position of equilibrium favors products. The larger K_c, the farther the reaction is toward completion.

Determining K_c from Other Reactions

Remember from the thermodynamics module that values for quantities such as ∆H could be obtained by manipulation of chemical reactions (according to Hess' Law). Equilibrium constants can be determined in a similar manner, however there are some significant differences in the way the K_c values for individual reactions are changed or combined.

In thermodynamics, when a reaction is reversed, the sign of the thermodynamic quantity is changed. For example:

2 Cu(s) + O₂ (g) → 2 CuO(s) ∆H^o = -310 kJ

2 CuO(s) → 2 Cu(s) + O₂ (g) ∆H^o = +310 kJ

When reactions are reversed, the value of the equilibrium constant, K_c, changes in a different way.

Look at the Kc for the following reaction.
aA + bB → cC + dD
Kc=[C]^c[D]^d /[A]^a[B]^b
Now, reverse the reaction and watch what happens to Kc.
cC + dD aA + bB
Kc=A]^a[B]^b/[C]^c[D]^d

Can you see that K_c flipped over? So, when an equilibrium reaction is reversed, K_c becomes the reciprocal, .

In thermodynamics, when reactions are multiplied by a factor, recall what happens to the thermodynamic quantity.

Answer: (The thermodynamic quantity, like ΔH, is also multiplied by that factor)

Look at what happens to K_c when a reaction is multiplied by a factor. To make this easier to see, the coefficients of 1 have been used (and written) in the equation. Normally we do not write 1.

1A + 1B 1C + 1D
Kc= [C]'[D]' [A]'[B]'
Now we will multiply the reaction by a factor of 2. Watch how Kc changes.
2A + 2B 2C + 2D
K = [C]²[D]²/[A]²[B]²
We see here that the value of Kc became Kc².

When an equilibrium reaction is multiplied by a factor, K_c is raised to that power.

In thermodynamics, when reactions are combined, their thermodynamic values are added together Look at this example to see what happens to Kc when reactions are combined.
Reaction 1 A + B C + D
Kc=[C][D]/[A][B]
Reaction 2 A 2B + D
K= [B]²[D]/[A]
Reaction 3 2A B+C+ 2D
Kc=[B][C][D]²/[A]²
Can you tell the relationship between Kc for the first and second reactions and the combination reaction? They were multiplied.
Kc=[C][D]/[A][B] x K [B]²[D]/[A] =
[C][D][B]²[D]/[A][B][A]= [B][C][D]²/[A]² .

K_c and K_p

When all the reactants and products of an equilibrium system are gases, you can use either concentrations or pressures to write the equilibrium law. For example,

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

$LaTeX: K_c=\frac{[NH_3]^2}{[NH_2][H_2]^3}$

When it is written in terms of pressure, we call the equilibrium constant K_P.

$LaTeX: K_P=\frac{{P_{NH_3}}^2}{{P_{(N_2)}P_{(H_2)}}^3}$

The value of K_p will be different from K_c. The ideal gas law can be rearranged to derive a formula to convert between K_c and K_p.

Kp =K (RT)^Δη(g)
R = 0.0821 Latm/molk
Δmoles of gas products minus moles of gas reactants

Example
At 500°C, this reaction: N2(g) + 3H2(g) E 2NH3(g) K. = 6.0 X 10.2
a) Write the equilibrium law using both pressure and concentration. b) What is the value of Kp?
K₁ = [NH,]/[N₂][H₂] = 6.0 X 102
P. NH₃²/ PN₃PH₂³=?
K = K(RT) = (6.0 X 10^-2)[(0.0821 L atm/mol K) (773K)]^(2-4)
K = (6.0 X 102)(63.5)² = 1.5X10^-5

Remember to work on the module practice problems as you complete each section of content.

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