KAE_The Meaning of Kc Lesson

The Meaning of Kc

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 Kc, 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]}}Kc=[products][reactants]

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

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

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

Kc is very small

Extremely small amount of products formed

Position of equilibrium favors reactants

Kc≈1

Position of equilibrium lays midway toward completion

Kc is very large

Very large amount of products formed

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

Determining Kc 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 Kc 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)               ∆Ho = -310 kJ

2 CuO(s) → 2 Cu(s) +   O2 (g)             ∆Ho = +310 kJ

When reactions are reversed, the value of the equilibrium constant, Kc, 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 Kc flipped over? So, when an equilibrium reaction is reversed, Kc 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 Kc 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, Kc is raised to that power.

In thermodynamics, when reactions are combined, their thermodynamic values are added togetherLook 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.

Kc and Kp

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}Kc=[NH3]2[NH2][H2]3

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

LaTeX: K_P=\frac{{P_{NH_3}}^2}{{P_{(N_2)}P_{(H_2)}}^3}KP=PNH32P(N2)P(H2)3

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

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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