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Balanced Equations And Equilibrium Constants

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    1365

    1. Introduction
    2. Rules to Keep in Mind
    3. Problems
    4. Solutions
    5. 1) SnO2(s) + 2 H2(g) <→ Sn(s) + 2 H2O(g)
    6. References
    7. Contributors

    Introduction

    To create the basic equilbrium constant, K, the chemical equation must first be balanced. Remember, nothing is lost or gained in a chemical reaction. The atoms are just bonded differently and create separate compounds with the input or release of chemical energy.

    Consider an element 'X':,

    #atoms for X on reactants = # atoms for X on products

    Another general example:

    (A2B)2 ---> 4A + B2

    Both the #atoms of A and B are equal on both sides.

    Example

    Find the equilibrium constant for the following reaction:

    \[ (NH_4)_2CO_3\,(s) \rightarrow NH_3 \,(g) + CO_2\, (g) + H_2O\,(g) \]

    Solution

    Notice: The reaction is not balanced. You need to first balance the equation before writing out the equilibrium constant.

    Part 1: Balancing the Reaction.

    Step 1: Make a chart with the initial number of atoms on the reactants and products side for each element:

    Element #atoms as reactants #atoms as products
    N 2 1
    H 8 5
    C 1 1
    O 3 3

    Step 2: Balance the Nitrogen:

    \[ (NH_4)_2CO_3(s) \rightarrow 2 NH_3(g)+ CO_2(g)+ H_2O(g) \]

    Step 3: Check to see if the equation is balanced:

    Element #atoms as reactants #atoms as products
    N 2 2
    H 8 8
    C 1 1
    O 3 3

    There are the same number of atoms of each element on the reactants side as there are on the products side. The equation is balanced!

    Part 2: Find the Equilibrium Constant.

    For the general equation, aA + Bb ---> cC + dD,

    As you can see, the equilibrium constant is essentially the ratio of the concentration of the products over the concentration of the reactants to the power of each of their respective coefficients. We use the chemical equilibrium to determine the extent and drive of the chemical reaction. Recall that when solving for equilibrium constants, the activities of pure solids and liquids are one, so (NH4)2CO3(s) is not included in the equation.

    Kc = [NH3]2[CO2][H2O]

    It is more appropriate to use Kp because all the constituents in the equation are gases:

    Kp = (PNH3)2(PCO2)(PH2O)

    Side note: Remember, standard pressure is in atmospheres, atm.

    What's the relationship between Kc and Kp?

    Kp = Kc (RT)(change in n)

    change in n/'delta' n = (#moles of products- #moles reactants)

    R= .08206L-atm/mol-K

    T= temperature in Kelvin

    Rules to Keep in Mind

    1) When two balanced equations are added together, their equilibrium constants are multiplied to give the final equilibrium constant of the added reaction.

    Example:

    \[ N_2(g) + (1/2)O_2(g) \rightleftharpoons N_2O(g) \;\; K_1= 5.4 \times 10^{-19} \]

    \[ N_2(g) + O_2(g) \rightleftharpoons 2NO(g)\;\; K_2= 4.6 \times 10^{-31} \]

    Given overall reaction (need to REVERSE first reaction, see Rule 2 for how this changes the equilibrium constant):

    N2O(g) + (1/2)O2(g) <---> 2NO(g) K = (1/K1)*(K2) = 8.5*10-13

    2) When we reverse an equation we invert the value of K. Kreverse reaction = 1/(Kforward reaction)

    3) When we multiply coefficients of a balanced equation by a common factor, we raise the equilibrium constant to that power.

    A<--->P equilibrium constant = K

    Let c = a constant (1, 2, 3, ...)

    cA<--->cP equilibrium constant = Kc

    1. When we divide the coefficients in a balanced equation by a common factor, we take the corresponding root of the equilibrium constant. (Similar to Rule 3; let c=the number you are dividing by. The overall equation would be multiplied by 1/c, so therefore the equilibrium constant would be raised to 1/c or the root of c).
    2. The equilibrium constant is unitless!
    3. Again, equilibrium constants do not contain pure solid and liquid concentrations because their partial pressures/concentrations do not change in a chemical reaction and their activities are equal to 1.
    4. A reaction is considered to go to completion if K>1010 or is not occurring in the forward direction if K<10-10 This is reasonable because if the product concentrations/partial pressures are >> the reactants concentrations/partial pressures at equilibrium, the equilibrium constant is very large, and the reaction is likely proceeding in the forward direction. The opposite is true when the products concentrations/partial pressures at equilibrium are much less than the reactants concentrations/partial pressures at equilibrium.
    5. Last, but not least, remember to always balance the chemical equation before calculating your equilibrium constant!

    Problems

    Balance the following reactions. Then, compute the appropriate equilibrium constants.

    1) SnO2(s) + H2(g) <→ Sn(s) + H2O(g)

    2) Mg(OH)2(aq) + H3PO4(aq) <--> Mg3(PO4)2(aq) + 4H2O(l)

    3) N2(g) + H2(g) <--> NH3(g)

    Solutions

    1) SnO2(s) + 2 H2(g) <→ Sn(s) + 2 H2O(g)

    Kp = (PH2O)2/(PH2)2

    2) 3Mg(OH)2(aq) + 2H3PO4(aq) <--> Mg3(PO4)2(aq) + 6H2O(l)

    Kc = [Mg3(PO3)2]/([Mg(OH)2]3[H3PO4]2)

    3) N2(g) + 3H2(g) <--> 2NH3(g)

    Kp = (PNH3)2/((PN2)(PH2)3)

    References

    1. Petrucci, Ralph H., et al. General Chemistry: Principles & Modern Applications. 9th ed. Upper Saddle River, New Jersey 2007.

    Contributors

    • Charlotte Hutton, Wen Yu,Mahtab Danai (UCD)