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Summary 11 - Equilibrium II - Notes

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Detailed and comprehensive notes for Topic 11 - Equilibrium II for A-Level Edexcel Chemistry. Based directly off the spec points, so you know you're not missing any content.

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Topic 11 – Equilibrium II

In order to develop their practical skills, students should be encouraged to carry out a range of practical experiments related to this topic. Possible
experiments include determining the value for an equilibrium constant for a simple esterification reaction. Mathematical skills that could be
developed in this topic include constructing expressions for Kc and Kp and calculating values with relevant units, estimating the change to the value of
an equilibrium constant when a variable changes. Within this topic, students can consider how chemists can use the concept of equilibria to predict
quantitatively the direction and extent of chemical change.


1. be able to deduce an expression for Kp, for homogeneous and heterogeneous systems, in terms of
equilibrium partial pressures in atm

Give an expression for Kp for the following reaction: 2 S O 2 ( g )+O 2 ( g) ⇌ 2 S O 3( g ). State the unit.


p ( S O 3 )2
K p= , atm-1
p ( S O 2 )2 × p ( O 2 )

Give an expression for Kp for the following reaction: N H 4 H S (s ) ⇌ N H 3 ( g )+ H 2 S ( g) . State the unit.

K p =p ( H 2 S ) × p ( N H 3 ) , atm 2




2. be able to calculate a value, with units where appropriate, for the equilibrium constant (K c and Kp) for
homogeneous and heterogeneous reactions, from experimental data

State the difference between a homogenous and a heterogenous reaction.

A homogenous reaction is one where the reactants and products are all in the same state.

A heterogenous reaction is one where the reactants and products are not in the same state.


2 moles of PCl5 vapour are heated to a temperature T1 in
a vessel of volume 2.0 x 101 dm3. The equilibrium mixture
contains 1.2 moles of chlorine. Calculate Kc for the
equilibrium at temperature T1.


PC l 5 ( g ) ⇌ PC l 3( g )+C l 2( g )

PCl5 PCl3 Cl2

I 2 0 0
C -1.2 +1.2 +1.2
E 0.8 1.2 1.2

[ PC l3 ][ C l2 ] ( 1.2÷ 2 ) × ( 1.2÷ 2 )
K c= =
[PC l 5 ] ( 0.8 ÷2 )

−3
K c =0.09mol d m

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