Calculating Equilibrium Constant (ICE Table)

 

This is part of the HSC Chemistry course under the topic Calculating Equilibrium Constant. 

HSC Chemistry Syllabus

  • Deduce the equilibrium expression (in terms of Keq) for homogeneous reactions occurring in solution. 
  • Perform calculations to find the value of Keq and concentrations of substances within an equilibrium system, and use these values to make predictions on the direction in which a reaction may proceed. 
  • Qualitatively analyse the effect of temperature on the value of Keq
  • Conduct an investigation to determine Keq of a chemical equilibrium system, for example:

– Keq of the iron (III) thiocyanate equilibrium

    How to Calculate Equilibrium Constant

    In this video, we will be exploring how to calculate the equilibrium constant and look at what the formula for providing the equilibrium expression is. 

    What is Equilibrium Constant

    • The equilibrium constant (Keq) represents the nature of a reaction at dynamic equilibrium. The value describes the relative quantity or concentration of reactants and products when the forward and backward reaction rates equal. 
    • For a general equation involving reactants A, B and products C, D in aqueous form:

    $$aA + bB \rightleftharpoons cC + dD$$

     

    • The equilibrium constant is:

     $$K_{eq}=\frac{{[C]}^c{[D]}^d}{{[A]}^a{[B]}^b}$$

     

    • Keq must always be positive (Keq > 0).

    • A large Keq means there are more products present compared to reactants at equilibrium. This does not mean the forward reaction rate is greater than the backward reaction rate because the two are equal at equilibrium.
    • A small Keq means there are more reactants present compared to products at chemical equilibrium. 
    • Only concentration/partial pressure of chemical species in aqueous and gas forms can be written in the equilibrium expression.
    • For example, the dynamic equilibrium involving iron (III) thiocyanate’s equilibrium:

    $${Fe^{3+}}_{(aq)} + \; {SCN^-}_{(aq)} \rightleftharpoons \; {FeSCN^{2+}}_{(aq)} \;\;\; \Delta{H}<0$$

     

     $$K_{eq}=\frac{[FeSCN^{2+}]}{[Fe^{3+}][SCN^–]}$$

     

    • Equilibrium constant of a reverse reaction is the reciprocal of that of the forward reaction:

    $${FeSCN^{2+}}_{(aq)} \rightleftharpoons {Fe^{3+}}_{(aq)} + \; {SCN^-}_{(aq)} \;\;\; \Delta{H}>0$$

     

     $$K_{eq(reverse)}=\frac{[Fe^{3+}][SCN^–]}{[FeSCN^{2+}]}=\frac{1}{K_{eq(forward)}}$$

     

    • Equilibrium constant changes when the stoichiometric ratio of a reaction changes. For example:

    $${2Fe^{3+}}_{(aq)} + \; {2SCN^-}_{(aq)} \rightleftharpoons \; {2FeSCN^{2+}}_{(aq)} \;\;\; \Delta{H}<0$$

     

    $$K_{eq(new)}=\frac{[FeSCN^{2+}]^2}{[Fe^{3+}]^2[SCN^–]^2}={(K_{eq})}^2$$

     

    Equilibrium Calculation

    Equilibrium calculation questions can be approached using the ‘ICE table’ method.

    I: initial reaction concentration

    C: change in reaction concentration

    E: equilibrium reaction concentration.

     

    Example 1

    A mixture of 5.00 mole H2(g) and 10.0 mole I2(g) are placed in a 5.00 L container at 450ºC and allowed to come to equilibrium. At equilibrium, the concentration of HI(g) is 1.87 M.

    (a) Calculate the value for the equilibrium constant, Kc, for this reaction under these conditions.

    $$H_{2(g)}+I_{2(g)} \rightleftharpoons 2HI_{(g)}$$

     

    		 `[H_(2(g))]` `[I_(2(g)]` `[HI_((g))]`

    Initial

    		 `5.00/5.00=1.00 M` `10.0/5.00=2.00 M`

    0

    Change

    		 `-x` `-x` `+2x`

    Equilibrium

    		 `1.00-x` `2.00-x`

    1.87

     

    $$0+2x=1.87 M$$

     

    $$x=0.935 M$$

     

    $$K_{eq}=\frac{{[C]}^c{[D]}^d}{{[A]}^a{[B]}^b}$$

     

    $$K_{eq}=\frac{{[HI]}^2}{[H_{2}][I_{2}]}$$

     

    $$K_{eq}=\frac{{(1.87)}^2}{(1.0-0.935)(2.0-0.935)}$$

     

    $$K_{eq}=50.5 (3 s.f.)$$ 

     

    (b) What does the equilibrium constant inform us about the relative concentration of reactants and products at equilibrium?

    The equilibrium constant is large (50.5) which means there are much more products than reactants at equilibrium.

    (c) Using your answer in part (a), calculate the equilibrium constant for the reaction:

    $$\frac{1}{2}H_{2(g)}+\frac{1}{2}I_{2(g)} \rightleftharpoons HI_{(g)}$$

     

    Previous section:  Practical Investigations of Le Chatelier's Principle (Combustion of Magnesium and Burning of Steel Wool)

    Next section: Calculating Reaction Quotient, Effect of Temperature on Equilibrium 

     

    RETURN TO MODULE 5: EQUILIBRIUM AND ACID REACTIONS