Kc Relationship to Balanced Chemical Equation

There are times when we are not able to run a particular reaction in the lab to determine Kc. We might run two or more reactions that will sum to the overall equation and then multiply the values of Kc to determine the equilibrium constant for the overall reaction.

First, make sure you have a balanced equation. If an equation is reversed, take the reciprocal of Kc. If the coefficients in the equation are multiplied by a common factor, then raise the equilibrium constant to the corresponding power. If two or more equations are added, multiply the equilibrium constants–Koverall = K1 x K2 x K3 . . . or Qoverall = Q1 x Q2 x Q3 . . . .

The following reaction has Kc = 5.8 x 105 at 298 K.

N2 (g) + 3 H2 (g) ⇄ 2 NH3 (g)   (equation 1)

From the given equation above, calculate Kc for the reaction below, equation 2.

NH3 ⇄ 1/2 N2 + 3/2 H2   (equation 2)

We can see that equation 2 is the reverse of equation 1, which means we need to take the reciprocal of Kc. The coefficients in equation 2 have been divided by 2, therefore we take the square root of 1/Kc.

\(\displaystyle \Bigl(\frac{1}{5.8\times 10^5}\Bigr)^{1/2}\;=\;1.3\times 10^{-3}\)
 
Kc for equation 2 is equal to 1.3 x 10-3.

In another example, we need to manipulate two equations for them to sum to the overall equation. Here we will determine Kc for the following equation:

N2 (g) + O2 (g) + Br2 (g) ⇄ 2 NOBr (g)    (equation 3)

We will use the next two equations to calculate Kc.

NO (g) + 1/2 Br2 (g) ⇄ NOBr (g) Kc = 1.4    (equation 4)

2 NO (g) ⇄ N2 (g) + O2 (g) Kc = 2.1 x 1030    (equation 5)

The stoichiometric coefficients can be multiplied by some factor or the equation (s) can be reversed. We can multiply equation 4 by a factor of 2. For equation 5, we can reverse it because O2 and N2 are reactants in the overall equation.

\(\displaystyle \require{cancel} \cancel{2\;NO\;(g)}\;+\;Br_2(g)\;⇄\;2\;NOBr\;(g)\;\;\;\;\; K_c\;=\;(1.4)^2\)
\(\displaystyle N_2\;(g)\;+\;O_2\;(g)\;⇄\;\require{cancel} \cancel{2\;NO\;(g)}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;K_c\;=\;\frac{1}{2.1\times 10^{30}}\)
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\(\displaystyle N_2\;(g)\;+\;O_2\;(g)\;+\;Br_2\;(g)\;⇄\;2\;NOBr\;(g)\;\)
\(\displaystyle and\;K_c\;=\;(1.4)^2\times \frac{1}{2.1\times 10^{30}}\;=\;9.3\times 10^{-31}\)

Kc for equation 3 is 9.3 x 10-31.

Watch the Following Video

 
Worksheet: Kc and Kp

Exercises

Exercise 1. Given the following,

2 NO (g) + 2 H2 (g) ⇄ N2 (g) + 2 H2O (g)    Kc = 6.7 x 102

what is the equilibrium constant for the following reaction?

1/2 N2 (g) + H2O (g) ⇄ NO (g) + H2 (g)    K’c = ?

Check Answer/Solution to Exercise 1

Exercise 2. Using the following equations,

H+ (aq) + F (aq) ⇄ HF (aq)     Kc = 1.5 x 103
2 H+ (aq) + C2O4-2 (aq) ⇄ H2C2O4 (aq)    Kc = 2.6 x 105

to find the value of the equilibrium constant for the following reaction.

2 F (aq) + H2C2O4 (aq) ⇄ 2 HF (aq) + C2O42- (aq)   Kc = ?

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