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

Revision Note

Test Yourself
Sonny

Author

Sonny

Expertise

Chemistry

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 [O2increases

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

Worked example

Factors which increase Kvalue:

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 K
  • 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

Exam Tip

  • 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

You've read 0 of your 0 free revision notes

Get unlimited access

to absolutely everything:

  • Downloadable PDFs
  • Unlimited Revision Notes
  • Topic Questions
  • Past Papers
  • Model Answers
  • Videos (Maths and Science)

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

Sonny

Author: Sonny

Sonny graduated from Imperial College London with a first-class degree in Biomedical Engineering. Turning from engineering to education, he has now been a science tutor working in the UK for several years. Sonny enjoys sharing his passion for science and producing engaging educational materials that help students reach their goals.

Join over 500 thousand students
getting better grades