CAIE A-Level Chemistry 7.1 Chemical and Dynamic Equilibrium

Practise defining dynamic equilibrium and predicting how temperature, concentration, pressure or a catalyst affects a system.

Exam points

state that forward and reverse rates are equal in a closed system while concentrations remain constant
use Le Chatelier’s principle to predict the direction of an equilibrium shift after a condition changes
interpret yield, composition or colour evidence without confusing equilibrium position with reaction rate

Covered syllabus points

7.1.1—What is meant by a reversible reaction 7.1.2—Le Chatelier’s principle as: if a change 7.1.3—Le Chatelier’s principle to deduce (from 7.1.4—Expressions for equilibrium constants 7.1.5—The terms mole fraction and partial pressure 7.1.6—Expressions for equilibrium constants 7.1.7—The Kc and Kp expressions to carry out 7.1.8—The quantities present at equilibrium 7.1.9—Whether changes in temperature, concentration 7.1.10—Conditions used in the Haber process

[Maximum number: 11]

The elements sodium to chlorine, in the third period, all form oxides.

(a)

$\mathrm{SO}_{3}$ is produced by the reaction between $\mathrm{SO}_{2}$ and $\mathrm{O}_{2}$ in the Contact process. A dynamic equilibrium is established.

2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}) \quad \Delta H=-196 \mathrm{~kJ} \mathrm{~mol}^{-1}

[ 5 ]

(i)

Explain why increasing the total pressure, at constant temperature, increases the rate of production of $\mathrm{SO}_{3}$ and increases the yield of $\mathrm{SO}_{3}$ .
rate
yield

The graph shows how the concentrations of all three species in the system change with time for a typical reaction mixture. The gradients of all three lines decrease with time and then level off in this dynamic equilibrium.

Figure for Question (i) — CAIE A-Level Chemistry AS

[ 4 ]

(ii)

Explain why all three lines become horizontal.

[ 1 ]

(b)

2.00 moles of $\mathrm{SO}_{2}(\mathrm{~g})$ and 2.00 moles of $\mathrm{O}_{2}(\mathrm{~g})$ are sealed in a container with a suitable catalyst, at constant temperature and pressure. The resulting equilibrium mixture contains 1.98 moles of $\mathrm{SO}_{3}(\mathrm{~g})$ .
The total volume of the equilibrium mixture is $40.0 \mathrm{dm}^{3}$ .

2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})

[ 6 ]

(i)

Write the expression for the equilibrium constant, $K_{\mathrm{c}}$ , for the reaction between $\mathrm{SO}_{2}(\mathrm{~g})$ and $\mathrm{O}_{2}(\mathrm{~g})$ to produce $\mathrm{SO}_{3}(\mathrm{~g})$ .

K_{c}=

[ 1 ]

$K_{c}=\frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{SO}_{2}\right]^{2} \times\left[\mathrm{O}_{2}\right]}$

(ii)

Calculate the amount, in moles, of $\mathrm{SO}_{2}(\mathrm{~g})$ and $\mathrm{O}_{2}(\mathrm{~g})$ in the equilibrium mixture.

\begin{array}{r} \mathrm{SO}_{2}(\mathrm{~g})=\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \mathrm{mol} \\ \mathrm{O}_{2}(\mathrm{~g})=\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \mathrm{mol} \end{array}

[ 2 ]

$\mathrm{SO}_{2}=0.02(\mathrm{~mol})$ $\mathrm{O}_{2}=1.01(\mathrm{~mol})$

(iii)

Use your answers to (d)(i) and (d)(ii) to calculate the value of $K_{\mathrm{c}}$ for this equilibrium mixture. Give the units of $K_{\mathrm{c}}$ .

\begin{array}{r} K_{\mathrm{c}}= \\ \text { units }= \end{array}

[ 3 ]

$\begin{aligned} K_{c} & =\frac{(1.98 / 40)^{2}}{(0.02 / 40)^{2}(1.01 / 40)} & =3.88 / 3.882 \times 10^{5} \end{aligned}$

Units $=\mathrm{dm}^{3} \mathrm{~mol}^{-1} / \mathrm{mol}^{-1} \mathrm{dm}^{3}$

M1

[Maximum number: 7]

(a)

Define Le Chatelier's principle.

[ 2 ]

(b)

Reaction 1 describes the reversible reaction between yellow $\mathrm{Fe}^{3+}(\mathrm{aq})$ and colourless $\mathrm{SCN}^{-}(\mathrm{aq})$ to produce red $\mathrm{FeSCN}^{2+}(\mathrm{aq})$ .
reaction 1

\underset{\text { yellow }}{\mathrm{Fe}^{3+}(\mathrm{aq})}+\underset{\text { colourless }}{\mathrm{SCN}^{-}(\mathrm{aq})} \rightleftharpoons \underset{\text { red }}{\mathrm{FeSCN}^{2+}(\mathrm{aq})} \quad \Delta H=-\mathrm{x} \mathrm{~kJ} \mathrm{~mol}^{-1}

A mixture of $\mathrm{Fe}^{3+}(\mathrm{aq}), \mathrm{SCN}^{-}(\mathrm{aq})$ and $\mathrm{FeSCN}^{2+}(\mathrm{aq})$ is at equilibrium at $20^{\circ} \mathrm{C}$ .
The temperature of this mixture is then increased to $50^{\circ} \mathrm{C}$ and allowed to reach equilibrium.
Deduce the changes that occur, if any, in the equilibrium mixture at $50^{\circ} \mathrm{C}$ compared to the equilibrium mixture at $20^{\circ} \mathrm{C}$ .
- change in appearance
- change in relative concentration of $\mathrm{FeSCN}^{2+}(\mathrm{aq})$
- change in value of the equilibrium constant, $K_{\mathrm{c}}$

[ 3 ]

M1 change in appearance paler red / more yellow / less orange

M2 change in relative concentration of $\mathrm{FeSCN}^{2+}(\mathrm{aq})$

lower

M3 change in value of the equilibrium constant, $K_{\mathrm{c}}$

lower

(c)

In another experiment, equimolar amounts of $\mathrm{Fe}^{3+}(\mathrm{aq})$ and $\mathrm{SCN}^{-}(\mathrm{aq})$ are mixed together and allowed to reach equilibrium. The total volume of the mixture is $25.0 \mathrm{~cm}^{3}$ .
reaction $1 \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{SCN}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{FeSCN}^{2+}(\mathrm{aq})$
At equilibrium the mixture contains:
- $\left[\mathrm{SCN}^{-}\right]=1.30 \times 10^{-3} \mathrm{moldm}^{-3}$
- $\left[\mathrm{FeSCN}^{2+}\right]=0.300 \times 10^{-3} \mathrm{moldm}^{-3}$ .

[ 2 ]

(i)

Calculate $\mathrm{K}_{\mathrm{c}}$ for reaction 1 and state its units.

Show your working.

K_{\mathrm{c}}=

units

[ 2 ]

M1 value for $K_{\mathrm{c}}=178$

M2 units $=\mathrm{mol}^{-1} \mathrm{dm}^{3}$

[Maximum number: 8]

Hydrogen halides are compounds formed when halogens (Group 17 elements) react with hydrogen. The bond polarity of the hydrogen halides decreases from HF to HI.

Some relevant data are shown in the table.

Table for Question 7.1 Chemical equilibria and dynamic equilibrium question 3 — CAIE A-Level Chemistry AS

(a)

Hydrogen chloride undergoes a reversible reaction with oxygen.

4 \mathrm{HCl}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{Cl}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{~g})

The reaction is carried out at $400^{\circ} \mathrm{C}$ in the presence of a copper(II) chloride catalyst.

[ 8 ]

(i)

The reaction exists in dynamic equilibrium.

The reaction was repeated at $1000^{\circ} \mathrm{C}$ and the same pressure.
State and explain the effect on the composition of the equilibrium mixture of the change in temperature.

[ 2 ]

Reaction is exothermic.

Increased temperature shifts the equilibrium to the left AND decreases the yield of products $\mathrm{Cl_2}$ and/or $\mathrm{H_2O}$ , so less product is formed.

(ii)

When 1.60 mol of HCl are mixed in a sealed container with 0.500 mol of $\mathrm{O}_{2}$ at $400^{\circ} \mathrm{C}$ , 0.600 mol of $\mathrm{Cl}_{2}$ and 0.600 mol of $\mathrm{H}_{2} \mathrm{O}$ are formed.

The total pressure inside the container is $1.50 \times 10^{5} \mathrm{~Pa}$ .
- Calculate the amounts, in mol , of HCl and $\mathrm{O}_{2}$ in the equilibrium mixture.

\begin{aligned} \mathrm{HCl} & =\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \mathrm{mol} \\ \mathrm{O}_{2} & =\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . . \mathrm{mol} \end{aligned}

- Calculate the mole fraction of $\mathrm{Cl}_{2}$ and hence the partial pressure of $\mathrm{Cl}_{2}$ in the equilibrium mixture.
mole fraction of $\mathrm{Cl}_{2}=$

p_{\mathrm{Cl}_{2}}=\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . . P a

[ 3 ]

(iii)

In a separate experiment, an equilibrium reaction mixture was found to contain the four gases at the partial pressures shown in the table.

Table for Question (iii) — CAIE A-Level Chemistry AS

K_{\mathrm{p}}=\frac{\left(p_{\mathrm{Cl}_{2}}\right)^{2} \times\left(p_{\mathrm{H}_{2} \mathrm{O}}\right)^{2}}{\left(p_{\mathrm{HCl}}\right)^{4} \times p_{\mathrm{O}_{2}}}

Use this information and the expression given for $K_{p}$ to calculate a value for $K_{p}$ . State the units of $K_{\mathrm{p}}$ .

\begin{array}{r} K_{\mathrm{p}}= \\ \text { units }= \end{array}

[ 2 ]

$K_p=\frac{(3.6\times10^4)^2\times(3.6\times10^4)^2}{(4.8\times10^4)^4\times3.0\times10^4}=1.05\times10^{-5}$
Units $=\mathrm{Pa^{-1}}$ .

(iv)

The reaction is repeated without a catalyst.

State the effect of this on $K_{\mathrm{p}}$ .

[ 1 ]

$K_p$ would not change.

All question bank results loaded

Covered syllabus points

7.1 Chemical equilibria and dynamic equilibrium question 1

Question (a)

Question (i)

Question (ii)

Question (b)

Question (i)

Question (ii)

Question (iii)

7.1 Chemical equilibria and dynamic equilibrium question 2

Question (a)

Question (b)

Question (c)

Question (i)

7.1 Chemical equilibria and dynamic equilibrium question 3

Question (a)

Question (i)

Question (ii)

Question (iii)

Question (iv)