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ChemWiki: The Dynamic Chemistry E-textbook > Physical Chemistry > Equilibria > Chemical Equilibria > The Equilibrium Constant > Balanced Equations And Equilibrium Constants

Balanced Equations And Equilibrium Constants

In a Balanced Equation, the total number of atoms of each element present is the same on both sides of the equation. Stoichiometric Coefficients are the coefficients required to balance a chemical equation. These are important because they relate the amounts of reactants used and products formed. The coefficients relate to the equilibrium constants because they are used to calculate them. For this reason, it is important to understand how to balance an equation before using the equation to calculate equilibrium constants.

Introduction

Here are some important concepts to remember for balancing an equation:

  1. An equation can be balanced only by adjusting the coefficients.
  2. The equation must include only the reactants and products that participate in the reaction.
  3. Never change the equation in order to balance it.
  4. If an element occurs in only one compound on each side of the equation, try balancing this element first.
  5. When one element exists as a free element, balance this element last.

Example  

  1. \[H_2\; (g) + O_2 \; (g) \rightleftharpoons H_2O \; (l) \]
Because both elements are by themselves, you can look at either first. Let’s look at oxygen. There are two atoms on the left side and one on the right side.  Multiply the right by 2 H2(g) + 1 O2(g) 2 H2O(l)
Next, look at the hydrogen. There are 4 atoms on the right, and only 2 atoms on the left. Multiply the hydrogen on left by 2 2 H2 (g) + O2(g) ↔ 2 H2O(l)
Now check. Hydrogen: on left 2 x 2 = 4, on right 2 x 2= 4. Oxygen: on left: 1 x 2 = 2, on right  2 x 1 = 2 .  All atoms balance. Done!

2 H2(g) + O2(g) ↔ 2 H2O(l)

Example  

\(2. Al \; (s) + MnSO_4 \; (aq) \rightleftharpoons Al_2(SO_4)_3 + Mn ; (s) \)

 First, look at the SO4. There is one on the left side, but three on the right side. Add a coefficient of three to the left side.  Al(s) + 3 MnSO4(aq) ↔ Al2(SO4)3 + Mn(s)
 Next, look at the Mn. There is one on the right side, but now three on the left side. Add a coefficient of three on the right side.  Al(s) + 3 MnSO4(aq) ↔ Al2(SO4)3 + 3 Mn(s)
 Lastly, look at Al. There is one on the left side and two on the right side. Add a coefficient of two on the left side. Check you numbers. Now you are done.  2 Al(s) + 3 MnSO4(aq)  ↔ Al2(SO4)3 + 3 Mn(s)

Example  

\(3. P_4S_3 + KClO_3 \rightleftharpoons P_2O_5 + KCl + SO_2 \)

 This problem is more difficult. First, look at the P. There are four on the left side and two on the right. Add a coefficient of two to the right.  P4S3 + KClO3 ↔ 2 P2O5 + KCl + SO2
 Next, look at the S. There are three on the left and one on the right. Add a coefficient of three to the right side.  P4S3 + KClO3 ↔ 2 P2O5 + KCl + 3 SO2
 Now look at the O. You want the total on each side to be equal. Now you have three on the left and 16 on the right. If you add a coefficient of 16 to the KClO3 on the left and the KCl on the right you will keep the same number of K and Cl, but increase the O.  P4S3 + 16 KClO3 <---> 2 P2O5 + 16 KCl + 3 SO2
 If you multiply the three others (P4S3, P2O5, and SO2) you will have equal numbers.  3 P4S3 + 16 KClO3 ↔ 2 (3) P2O5 + 16 KCl + 3 (3) SO2
Now check you numbers.  3 P4S3 + 16 KClO3 ↔ 6 P2O5 + 16 KCl + 9 SO2


Chemical Equlibrium

Now that we understand how to balance the equations, we can move on to using the equaions to find the equilibrium. Chemical Equilibrium is the state in which the reactants and products have no net change over time. This is when the forward and reverse reactions occur at equal rates. To determine the amount of each compound that will be present at equilibrium you must know the Equilibrium Constant. To determine the equilibrium constant you must consider the Chemical Reaction written in the form:

\( aA + bB \rightleftharpoons cC + dD\)

Using this standard equation, the Equilibrium Constant is defined as:

\( K_c = \dfrac{[C]^c * [D]^d}{[A]^a * [B]^b} \)

The Products are in the numerator and those of the reactants are in the denominator. The equation is a function of the concentrations, the upper case letters are the molar concentrations of the reactants and products (A, B etc.). The lower case letters are the stoichiometric coefficients that balance the equation.

An important thing to note about this equation is that pure liquids and solids are not used. This is because they are equal to one, so plugging the into the equation would have no effect on it. Although this many seem confusing, it is simiply because they would not change the equilibrium. They wouldn't have an effect on it due to the fact that no matter how much more you add, the reaction will only disolve as much as the solubility allows. For example, if you added more sugar to water after the equilibrium has been reached, the extra sugar will not disolve (assuming you do not introduce heat, which would increase the solubility). Because adding more will not change the equilibrium, you would say that it doesn't have an effect on it, hense you would leave it out of the calculations. You may have noticed this very thing taking place, and thinking about it can help you to better understand the concept.

K is related to to the Balenced Chemical Reaction

The following are some concepts that hold true when making sure the expression for K matches the corresponding balanced equation:

  • When we reverse an equation, we invert the value of K
  • When we multiply the coefficients in a balanced equation by a common factor, we raise the equilibrium constant to the corresponding factor.
  • When we divide the coefficients in a balanced equatoin by a common factor, we take the corresponding root of the equilibrium constant.
  • When individual equations are combined, their equilibrium constants are multiplied to obtain the equilibrium constant for the overall reaction.

Having a balanced equation is very important for using the constant because the coefficients become the powers of the products and reactants which they are reffering to. If the equation is not balanced, then the constants will be not be right.

K IS ALSO RELATED TO THE BALANCED CHEMICAL EQUATION OF GASES
For equilibiria in the gas phase, the equation is a function of the reactants and products Partial Pressures. The Equilibrium Constant is expressed as:

\( K_p = \dfrac{P_C^c * P_D^d}{P_A^a * P_B^b} \)

P represents partial pressure, usually in atmospheres. Here we must also remember that pure solids and liquids are not accounted for in the equation. Kc and Kp are related by the equation:

\( K_p = K_c(RT)^{\Delta n} \)

This equation relates the Balanced Equation to Chemical Equilibrium

\( \Delta n = (c+d) - (a+b) \)

This represents the change in gas molecules. a,b,c and d are the Stoichiometric Coefficients of the gas molecules found in the balanced equation.

Note: Both Kc and Kp do not have units. This is because they are direct comparisons of the products and reactants. Their units cancel out when doing the calculation and then when they are used in other calculations, not having a unit prevents further problems.

Example  

1. Using Kc

\( PbI_2 \rightleftharpoons Pb \; (aq) + I \; (aq) \)

First, balance the equation.

Check the Pb. There is one on each side, so we can leave it alone for now. Next check the I. There are two on the left side and one on the right side. To fix this, we add a coefficient of two to the right side. PbI2 ↔ Pb2+(aq) + 2 I-(aq)
Now check to make sure the numbers are equal. PbI2 ↔ Pb2+(aq) + 2 I-(aq)

Next, we can find Kc. Use these concentrations: Pb- 0.3 mol/L, I- 0.2 mol/L, PbI2- 0.5 mol/L

\( Kc = \dfrac{(0.3) * (0.2)^2}{(0.5)} \)

Kc= 0.024

Note: If the equation had not been balanced when the equilibrium constant was calculated, the concentration of I - would not have been squared. This would have given you an incorrect answer.

Example  

\( SO_2 \; (g) + O_2 \; (g) \rightleftharpoons SO_3 \; (g) \)

 First, make sure the equation is balanced.

Check to make sure S is equal on both sides. We see that there is one on each side. Next look at the O. There are four on the left side and three on the right. Your first thought might be to add a coefficient to the O2 on the left, but this will not work because you will have to increase the S on right as you increase the O. Instead, add a coefficient to the SO2 on the left and the SO3 on the right. 2 SO2 + O2 ↔ 2 SO3
Now check your numbers. As you can see the equation is now balanced. 2 SO2 + O2 ↔ 2 SO3

Now we can calculate Kp. If these are the partial pressures:  SO2- 0.25 atm, O2- 0.45 atm, SO3- 0.3 atm

\( Kp = \dfrac{(0.3)^2}{(0.25)^2 * (0.45)} \)

Kp= 3.2


Contributors

  • Charlotte Hutton, Sarah Reno, Curtis Kortemeier

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Last Modified
09:26, 2 Oct 2013

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