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Reaction Conditions & The Equilibrium Constant (Edexcel A Level Chemistry): Revision Note

Stewart Hird

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Temperature & the Equilibrium Constant

  • Changes in temperature change the equilibrium constants Kc and Kp

  • For an endothermic reaction such as:

   An increase in temperature:

   [H2] and [I2] increases

   [HI] decreases

   Because [H2] and [I2] are increasing and [HI] is decreasing, the equilibrium constant increases

  • For an exothermic reaction such as:

   An increase in temperature:

   [SO3] decreases

   [SO2] and [O2] increases

   Because [SO3] decreases and [SO2] and [O2] increases the equilibrium constant  decreases

Worked Example

Factors which increase Kp value:

What will increase the value of Kp of the following equilibrium?

2A (g) + B (g)  ⇌  2C (g)       ΔH = +6.5 kJ mol-1 

Answer

  • Only temperature changes permanently affect the value of Kp

    • An increase in temperature shifts the reaction in favour of the products.

    • The [ products ] increases and [ reactants ] decreases, therefore, the Kp value increases.

Temperature & the Equilibrium Position

  • How the equilibrium shifts with temperature changes:

Effects of temperature table, IGCSE & GCSE Chemistry revision notes

Effect on the value of the equilibrium constant

  • For a reaction that is exothermic in the forward direction, increasing the temperature pushes the equilibrium from right to left

  • Therefore, the value of the equilibrium constant will decrease as the ratio of [ products ] to [ reactants ] decreases

  • Conversely, if the temperature is raised in an endothermic reaction, the value of the equilibrium constant will increase

Changing Reaction Conditions

  • If all other conditions stay the same, the equilibrium constant Kc is not affected by any changes in concentration of the reactants or products

  • For example, the decomposition of hydrogen iodide:

2HI ⇌ H2 + I2

   The equilibrium expression is:

Changes that Affect the Equilibrium Constant equation 1

   Adding more HI makes the ratio of [ products ] to [ reactants ] smaller

To restore equilibrium, [H2] and [I2] increases and [HI] decreases

Equilibrium is restored when the ratio is 6.25 x 10-3 again

Changes in pressure

  • A change in pressure only changes the position of the equilibrium

  • If all other conditions stay the same, the equilibrium constant Kc is not affected by any changes in pressure of the reactants and products

  • The value of Kp is not affected by any changes in pressure.

  • Changes in pressure cause a shift in the position of equilibrium to a new position which restores the value of Kp 

  • This is analogous to what happens to Kc when you change concentration in an aqueous equilibrium; a shift restores equilibrium to a new position maintaining Kc

Presence of a catalyst

  • If all other conditions stay the same, the equilibrium constants Kp and Kc are not affected by the presence of a catalyst

  • A catalyst speeds up both the forward and reverse reactions at the same rate so the ratio of  [ products ] to [ reactants ] remains unchanged

  • Catalysts only cause a reaction to reach equilibrium faster

  • Catalysts, therefore, have no effect on the position of the equilibrium once this is reached

Worked Example

Hydrogen iodide is formed in the gas phase by the reaction of hydrogen and iodine:

H2 (g) + I2 (g)rightwards harpoon over leftwards harpoon2HI (g) 

The equilibrium constants at two different temperatures are related by the following expression:

ln open square brackets K subscript 2 over K subscript 1 close square brackets equals fraction numerator increment H over denominator R end fraction open square brackets 1 over T subscript 1 minus 1 over T subscript 2 close square brackets

At 763 K, the equilibrium constant K1 is 45.9. The enthalpy change for the reaction is ΔH = -26 500 J mol-1.

Calculate the value of the equilibrium constant K2 for this reaction at 718 K.  [Use the value of R= 8.31 J mol-1 K-1]

Answer:

Error converting from MathML to accessible text.

begin mathsize 14px style ln stretchy left square bracket fraction numerator K subscript 2 over denominator 45.9 end fraction stretchy right square bracket equals fraction numerator negative sign 26 space 500 over denominator 8.31 end fraction stretchy left square bracket 1 over 763 minus sign 1 over 718 stretchy right square bracket end style

ln stretchy left square bracket fraction numerator K subscript 2 over denominator 45.9 end fraction stretchy right square bracket equals 0.26194

fraction numerator K subscript italic 2 over denominator 45.9 end fraction equals space e to the power of 0.26194 end exponent

K subscript 2 space equals space 45.9 space cross times space 1.2995 space equals 59.64

  • This is an exothermic reaction, so you would expect a decrease temperature to cause the equilibrium to shift to the right. This is borne out by seeing an increase in the value of K as the temperature decreases

Examiner Tips and Tricks

  • In exams, you do not need to know the equation in this worked example

  • You will be given this equation if you are expected to work with it

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    Stewart Hird

    Author: Stewart Hird

    Expertise: Chemistry Lead

    Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.