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Chemical Equilibrium?

1. If you have this equilibrium reaction: 2HI (g)  H2 (g) + I2 (g)

What will happen to the equilibrium add more I2 or oxygen (O2)?

2. If you have a this equation: PCl5(g) <===> PCl3(g) + Cl2(g).

There is .0220 M PCl3 that you start with. And then at equilibrium at 15. liters you have .0333 PCL5.

So how many moles pcl5 are there?

How many mole/liters are there for pcl5?

How to do calculate Kc and Kp for this reaction???

I know that you have to use an ice table and Pv=nRT in to solve this…can you please help me? I have a test on this material tomorrow. Thank you.

Chemical Equilibrium
Reversible Reactions

Equilibrium

Factors Affecting An Equilibrium

Change in Concentration

Change in Pressure

Change in Temperature

The Effect of Catalysis on an Equilibrium

Equilibrium and Industrial Reactions

The Haber Process

The Contact Process

Equilibrium in Water

The pH Scale

Diluting Acids and Alkalis

The Continuous pH Scale

Equilibrium in Acids and Alkalis

Equilibrium in Salts

——————————————————————————–

Reversible Reactions
Many reactions in chemistry go 100% to completion e.g. burning methane, neutralisation. However, there are reactions which appear to be incomplete because they are reversible – the products collide and react to produce the original reactants.

Example 1: When ammonium chloride is heated, it decomposes to form ammonia and hydrochloric acid. When cooled, the gases combine to form ammonium chloride again.

NH4+Cl-(s) ® NH3(g) + HCl(g)

then:

NH3(g) + HCl(g) ® NH4+Cl-(s)

written as:

NH4+Cl-(s) Û NH3(g) + HCl(g)

Example 2: When pink cobalt chloride crystals are heated, they turn blue and steam is given off. When water is added to the blue crystals, they revert to their original pink colour.

Co2+(Cl-)2.6H2O(s) ® Co2+(Cl-)2(s) + 6H2O(l)

then:

Co2+(Cl-)2(s) + 6H2O(l) ® Co2+(Cl-)2.6H2O(s)

written as:

Co2+(Cl-)2.6H2O(s) Û Co2+(Cl-)2(s) + 6H2O(l)

This is the reaction that occurs, provided that the 6H2O does not escape, i.e. it is a closed system.

Equilibrium
Consider starting a reversible reaction with 100% of A and B. Initially the forwards rate is at a maximum because the concentration of the ‘reactants’ is at its maximum while the reverse reaction rate is zero because the concentration of the ‘products’, C and D, is virtually zero.

After some minutes, the concentration of the ‘reactants’, A and B, is lower, so the forwards rate is lower, while the concentration of the ‘products’, C and D, is greater, so the reverse rate is greater. Clearly, the two opposing rates will become equal.

************

To the observer, the quantity of rectants becoming products is the same quantity of products becoming reactants, i.e. the proportions of reactants and products is constant, no further change in proportions occurs.

N.B. The system has not stopped, i.e. it is a dynamic equilibrium. Both of the reactions occur simultaneously, not alternately.

N.B. The actual proportion of reactants to products is not necessarily 50% : 50%.

The same proportion of reactants and products, i.e. equilibrium position, is reached irrespective of how the equilibrium is started i.e. whether started as 100% reactants or 100% products or in any other proportion.

Experiment: Partition of Iodine Between Two Immiscible Solvents

Iodine in aqueous K+I- is a brown solution and iodine in hexane is a pink, non-aqueous solution.

*********

I2 (in K+I-(aq)) Û I2 (in hexane)

Experiment: Showing the Dynamic Nature of Chemical Equilibrium

****************

The system is an equilibrium between dissolution and precipitation, i.e.

dissolution®

C6H12O6(s)
Û
C6H12O6(aq)
¬precipitation

By using radioactive, 14C labelled, glucose, very quickly the radioactivity is detected in the solution. If the equilibrium were static, it should reain in the extra added 14C glucose.

Factors Affecting An Equilibrium Position

1. Change in Concentration
a) Increasing the concentration of a reactant shifts an equilibrium to the products (or right hand) side because the rate of the forwards reaction is increased.

b) Increasing the concentration of a product shifts an equilibrium to the reactant (or left hand) side because the rate of the reverse reaction is speeded up.

c) Decreasing the concentration of a reactant (by removal or by compounding it with something else or by precipitation) shifts an equilibrium to the reactants (or left hand) side because the forwards reaction is slowed down. The reverse reaction will ‘overtake’ the forwards reaction.

d) Decreasing the concentration of a product shifts an equilibrium to the products (or right hand) side because the reverse reaction is slowed and the forwards reaction ‘overtakes’.

2. Change in Pressure
Consider:

N2O4(g) Û
2NO2(g)
dinitrogen tetraoxide nitrogen dioxide
very light yellow, almost colourless brown

Increased pressure will cause more collisions to take place between gaseous molecules (pressure is a measure of the number of particles per unit volume).

an increase in pressure will favour the reaction involving the most particles
an increase in pressure favours 2NO2 molecules reacting more than one N2O4 reacting
the reverse reation is speeded up
the equilibrium shifts to the left hand side i.e. a lighter colour is produced (after an expected initial darkening caused by the original colour being ‘compressed’).
If the pressure is decreased (by expanding a syringe of gas, for example), the initial colour thins but rapidly darkens. The equilibrium has shifted to the right hand side because the reverse reaction has slowed down – its collisions have been reduced in frequency.

N.B. Pressure changes only matter if there is a different number of gas molecules on each side. Pressure changes are irrelevent if there are no gas molecules in the reaction, e.g.

H2(g) + I2(g)
Û
2HI(g)
2 moles of gas
2 moles of gas

Here there is no change in equilibrium position.

3. Change in Temperature
Generally, an increase in temperature speeds up a reaction, but DH values must be considered. An endothermic reaction is helped more by a temperature rise than an exothermic one, e.g.

endothermic®

N2O4(g) Û
2NO2(g)
dinitrogen tetraoxide ¬exothermic
nitrogen dioxide
very light yellow, almost colourless brown

increasing temperature will favour the forwards reaction more so than the reverse, and so the colour darkens and the equilibrium shifts to the right hand side
decreasing temperature slows the endothermic reaction the most, and so the forwards reaction slows drastically and the reverse reaction wins, which means that the colour lightens as the equilibrium shifts to the left hand side.

4. The Effect of Catalysis on an Equilibrium
**************

A catalyst provides an easier path for the reaction, the path for the reverse reaction is made equally easier. A catalyst will not shift an equilibrium position because both rates are equally increased. The equilibrium is achieved quicker in time and under easier conditions, however.

Equilibrium and Industrial Reactions

The Haber Process
N2 + 3H2 Û 2NH3

forwards reaction is exothermic

Conditions for best yield of ammonia:

a) Temperature – high temperature would favour the endothermic process, so if it’s too high, this could decompose too much of the ammonia. If it’s too low, the production would be too slow, or not enough ammonia would be produced – a compromise must be reached.

b) Pressure – increased pressure will produce more ammonia, but it is expensive in machinery and energy and it can also be dangerous. At 1 atmosphere pressure the yield is very low, and so high pressure is necessary.

c) Concentration – if the ammonia could be continuously removed, the equilibrium would shift to the right hand side. If extra nitrogen or hydrogen were added, the equilibrium would again shift to the right hand side.

d) Catalyst – the addition of a catalyst doesn’t increase the yield in any way, however, it does make the overall reaction proceed much quicker.

Solution: A temperature of 500°C, a pressure of 200 atmospheres, and an iron catalyst are used. The ammonium is removed by liquifying (condensing) and the unreacted nitrogen and hydrogen is continuously recycled. A yield of 15% is achieved, but it is quick and continuous and there is no wastage and so it costs relatively little. Equilibrium is never achieved.

The Contact Process
2SO2(g) + O2(g) Û 2SO3(g)

forwards reaction exothermic

Conditiions for best yield:

a) Temperature – a low temperature would hinder the reverse, endothermic, reaction, but the rate of production of SO3 would be too slow.

b) Pressure – increased pressure would help the forwards reaction as it involves more gas particles, but it is expensive and dangerous. Actually, at 1atmosphere pressure there is a 98% yield.

c) Concentration – a deliberate increase n SO2 and O2 would favour production of SO3 and the removal of SO3 would hinder the reverse reaction.

d) Catalyst – a good catalyst would make the whole system quicker and easier.

Solution: A temperature of 400°C, a pressure of 1 atmosphere and a vanadium pentoxide (V2O5(s)) catalyst are used. SO3 is removed by dissolving in concentrated sulphuric acid:

SO3 + H2SO4(l) ® Hsub>2S2O7(l)

H2S2O7 + H2O(l) ® (H4S2O8)2H2SO4(l)

i.e. 1 more H2SO4 is gained per SO3. Equilibrium is never achieved.

Equilibrium in Water
Pure water does conduct electricity very slightly.

H2O(l) Û H+(aq) + OH-(aq)

The equilibrium lies well to the left hand side (only 1 in 555 000 000 molecules ionise). The concentration of the two ions is equal, so it still has pH 7. The equilibrium can be upset, e.g. by reaction with calcium (or other MAZIT metal).

Ca(s) + 2H2O(l) ® Ca(OH)2(aq) + H2(g)

Ca(s) + 2H+(aq) ® Ca2+(aq) + H2(g)

This removes the H+(aq) from the water equilibrium, slowing the reverse reaction and leading to an excess of OH-(aq) being present and so the solution becomes alkaline.

The pH Scale
The pH scale is a measure of H+(aq) ion concentration written as [H+(aq)] (in mol l-1).

It is seen that when [H+(aq)]
=1 mol l-1 i.e. 1×100 mol l-1, pH=0
=0.1 mol l-1 i.e. 1×10-1 mol l-1, pH=1
=0.01 mol l-1

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