Topic: Equilibria

1:03 understand how the results of experiments involving the dilution of coloured solutions and diffusion of gases can be explained

Diffusion is the spreading out of particles in a gas or liquid. There is a net movement of particles from areas of high concentration to areas of low concentration until a uniform concentration is achieved.

 

i) dilution of coloured solutions

Dissolving potassium manganate(VII) in water demonstrates that the diffusion in liquids is very slow because there are only small gaps between the liquid particles into which other particles diffuse.

The random motion of particles cause the purple colour to eventually be evenly spread out throughout the water.

Adding more water to the solution causes the potassium manganate(VII) particles to spread out further apart therefore the solutions becomes less purple. This is called dilution.

 

ii) diffusion experiments

When ammonia gas and hydrogen chloride gas mix, they react together to form a white solid called ammonium chloride.

ammonia                  +              hydrogen chloride                 –>            ammonium chloride

NH3(g)                     +              HCl(g)                                     –>            NH4Cl(s)

A cotton wool pad was soaked in ammonia solution and another was soaked in hydrogen chloride solution. The two pads were then put into opposite ends of a dry glass tube at the same time.

The white ring of ammonium chloride forms closer to the hydrochloric acid end because ammonia particles are lighter than hydrogen chloride particles and therefore travel faster.

Even though these particles travel at several hundred metres per second, it takes about 5 min for the ring to form. This is because the particles move in random directions and will collide with air particles in the tube.

1:30 calculate percentage yield

Yield is how much product you get from a chemical reaction.

The theoretical yield is the amount of product that you would expect to get. This is calculated using reacting mass calculations.

In most chemical reactions, however, you rarely achieved your theoretical yield.

For example, in the following reaction:

     CaCO3     –>            CaO         +           CO2

You might expect to achieve a theoretical yield of 56 g of CaO from 100 g of CaCO3.

However, what if the actual yield is only 48 g of CaO.

By using the following formula, the % yield can be calculated:

    \[yield= \frac{actual\,amount\,of\,product}{theoretical\,amount\,of\,product} \]

    \[yield= \frac{48}{56} \]

    \[yield=0.86 \]

    \[\% yield=86\% \]

3:18 describe reversible reactions such as the dehydration of hydrated copper(II) sulfate and the effect of heat on ammonium chloride

Dehydration of copper(II) sulfate

 

Heating ammonium chloride

On heating, white solid ammonium chloride decomposes forming ammonia and hydrogen chloride gas. On cooling, ammonia and hydrogen chloride react to form a white solid of ammonium chloride:

 

3:20 (Triple only) know that the characteristics of a reaction at dynamic equilibrium are: the forward and reverse reactions occur at the same rate, and the concentrations of reactants and products remain constant

Features of a reaction mixture that is in dynamic equilibrium:

  1.   the concentrations of reactants and products remain constant
  2.   rate of forward reaction = rate of backward reaction

3:21 (Triple only) understand why a catalyst does not affect the position of equilibrium in a reversible reaction

A catalyst is a substance which increases the rate of reaction without being chemically changed at the end of the reaction.

A reversible reaction is one where the forward reaction and the backward reaction happen simultaneously. For example:

3H₂ + N₂ ⇋ 2NH₃

In such a reaction a catalyst speeds up both the forward and the backward reactions. Hence, although the system will reach dynamic equilibrium more quickly, the addition of a catalyst will not affect the position of equilibrium.

3:22 (Triple only) predict, with reasons, the effect of changing either pressure or temperature on the position of equilibrium in a reversible reaction (references to Le Chatelier’s principle are not required)

In a reversible reaction the position of the equilibrium (the relative amounts of reactants and products) is dependent on the temperature and pressure of the reactants.

If the conditions of an equilibrium reaction are changed, the reaction moves to counteract that change.

Therefore by altering the temperature or pressure the position of the equilibrium will change to give more or less products.

Adding a catalyst does not affect the position of the equilibrium.

 

Changing the temperature:

All reactions are exothermic in one direction and endothermic in the other way.

For this reaction the enthalpy change, ΔH is negative therefore the forward reaction is exothermic:

     CO(g)     +             2H2(g)                    ⇌            CH3OH(g)              ΔH = –91 kJ/mol

If temperature is decreased the position of the equilibrium will shift to the right because it is an exothermic reaction.

For this reaction the enthalpy change, ΔH is positive therefore the forward reaction is endothermic:

     CH4(g)                     +              H2O(g)                    ⇌            CO(g)      +              3H2(g)                     ΔH = +210 kJ mol–1

If temperature is increased the position of the equilibrium will shift to the right because it is an endothermic reaction.

Key point: an increase (or decrease) in temperature shifts the position of equilibrium in the direction of the endothermic (or exothermic) reaction

 

Changing the pressure:

Reactions may have more molecules of gas on one side than on the other.

For this reaction there are 2 molecules on the left and 4 molecules on the right:

     CH4(g)                     +              H2O(g)                    ⇌            CO(g)      +              3H2(g)                     ΔH = +210 kJ mol–1

If the pressure is increased the position of the equilibrium will shift to the left because there are fewer molecules on the left-hand side.

For this reaction there are 3 molecules on the left and 1 molecule on the right

     CO(g)     +             2H2(g)                    ⇌            CH3OH(g)              ΔH = –91 kJ/mol

If the pressure is decreased the position of the equilibrium will shift to the left because there are more molecules on the left-hand side.

Key point: an increase (or decrease) in pressure shifts the position of equilibrium in the direction that produces fewer (or more) moles of gas

Select a set of flashcards to study:

     Terminology

     Skills and equipment

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Section 1: Principles of chemistry

      a) States of matter

      b) Atoms

      c) Atomic structure

     d) Relative formula masses and molar volumes of gases

     e) Chemical formulae and chemical equations

     f) Ionic compounds

     g) Covalent substances

     h) Metallic crystals

     i) Electrolysis

 Section 2: Chemistry of the elements

     a) The Periodic Table

     b) Group 1 elements: lithium, sodium and potassium

     c) Group 7 elements: chlorine, bromine and iodine

     d) Oxygen and oxides

     e) Hydrogen and water

     f) Reactivity series

     g) Tests for ions and gases

Section 3: Organic chemistry

     a) Introduction

     b) Alkanes

     c) Alkenes

     d) Ethanol

Section 4: Physical chemistry

     a) Acids, alkalis and salts

     b) Energetics

     c) Rates of reaction

     d) Equilibria

Section 5: Chemistry in industry

     a) Extraction and uses of metals

     b) Crude oil

     c) Synthetic polymers

     d) The industrial manufacture of chemicals