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Shifting Equilibrium – The Effect of Pressure,Temperature and Concentration

Mr. Shields Regents Chemistry U13 L03

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Factors affecting EquilibriumFactors affecting Equilibrium

As we saw previously, there are several factors that canaffect an equilibrium:

1) Adding or removing HEAT2) Increasing PRESSURE3) Changing CONCENTRATION by adding, decreasing OR totally removing one or more reactants/products

1-3 will shift the equilibrium right or left but an equilibrium still exists

However, totally removing one of the reactants/products will destroyThe equilibrium and drive the reaction to completion

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PressurePressureWhen we discuss the effect of pressure on equilibriumReactions we are discussing reversible reactions in the GAS PHASE.

Rxn is in gas Phase:N2 (g) + 3H2 (g) ↔ 2NH3 (g) P has an Effect

Pressure has no effect on equilibrium reactions in theLiquid or solid phase. Rxn is not in the Gas Phase:

Pressure has no Effect

BaSO4 (s) + H2O ↔ Ba+2 (aq) + SO4-2 (aq)

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PressurePressureAccording to Le Chatelier’s Principle if PRESSURE is Increased the system will react to reduce the stress(i.e. it will try to reduce pressure)

In a gaseous equilibrium rxn it can do this by shiftingthe rxn in the direction that MINIMIZES the NUMBEROf MOLES OF GAS.

For example consider the following Rxn:

N2 (g) + 3H2 (g) ↔ 2NH3 (g) + 22KJ heat

How many moles of products and reactants are there?

4 moles 2 moles

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PressurePressureWhy would the system shift the equilibrium in the direction that decreases the # of moles of gas?

Remember Mole conversions & Gas laws?

What’s the volume of one mole of gas?

If I reduce the number of moles of gas what normally happens to the volume occupied by the gas?

But since our system has a fixed volume, as the # of particles decrease the Pressure has to decreases

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PressurePressure

N2 (g) + 3H2 (g) ↔ 2NH3 (g)

So let’s look at this reaction:

We would predict that the reaction would shift to theRIGHT when the pressure on the system increases.

And if the pressure decreases?

We would also predict that the reaction would shift LEFTWhen pressure on the system is decreased.

4 mol 2 mol

If I increase the pressure on the system what doesLe Chatelier predict & which way will the equilibrium shift?

WHY?

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The effect of P if The effect of P if Not EverythingNot Everything is a gas is a gas

NH4Cl (s) ↔ NH3 (g) + HCl(g)

We’ve considered what happens if there are gases onlyon both sides of the equilibrium equation.

But what happens if one of the products or reactants isnot a gas? For example …

In this case we simply consider that product or reactantTo represent 0 moles of gas! So… for the case above what is the effect of inc. & then dec. system pressure?

Increasing system pressure shifts the reaction left anddecreasing system pressure would shift it to the right.

0 moles 2 moles

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If the equilibrium equation has an equal number ofmoles of gas on both sides of the equation what happensIf we increase the pressure?

For example what happens to the following equilibriumif we increase the pressure from 2 atm to 4 atm?

An increases in pressure will have no effect on shifting theEqilibrium to either the right or left. Why?

A change in either direction will result in no change in the # ofmoles and thus no change in pressure.

H2 (g) + Cl2 (g) ↔ 2HCl (g)2 moles 2 moles

EFFECT OF PRESSURE WHEN # OF MOLES OF PRODUCT Gases EQUALS # OF MOLES OF REACTANT Gases

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Effect of TemperatureEffect of TemperatureTo understand the effect of temperature changes onequilibrium you 1st need to know whether the rxnIs Endothermic or Exothermic.

In all reversible rxns if one direction is ExothermicThe reverse rxn has to be Endothermic.

N2 (g) + 3H2 (g) ↔ 2NH3 (g) + 93KJ

exothermic

endothermic

For example…

Released heat

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Effect of TemperatureEffect of TemperatureIn an EXOTHERMIC rxn Le Chatelier’s Principle saysadding heat to the system will shift the rxn to the left

N2 (g) + 3H2 (g) ↔ 2NH3 (g) + Heat H -93KJ

And if in an Exothermic rxn if we remove heat from theSystem the Reaction will shift to the right

N2 (g) + 3H2 (g) ↔ 2NH3 (g) + Heat H -93KJ

Rxn shifted in this direction (const P)

Rxn shifted in this direction (const P)

Note: If is Positive (+) then reaction is Endothermic & Heat term Is on the left (reactant) side of the equation.

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Effect of TemperatureEffect of TemperatureIn an ENDOTHEMIC rxn Le Chatelier’s Principle Predictswe will get an effect opposite to an exothermic rxn

Heat + NH4CL (s) ↔ NH3 (g) + Cl2 (g)

Rxn shifted in this direction (const P)

Adding heatTo the system

Heat + NH4CL (s) ↔ NH3 (g) + Cl2 (g)

Rxn shifted in this direction (const P)

Removing heatFrom thesystem

Endothermic Rxn

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Effect of ConcentrationEffect of ConcentrationLet’s now see what happens when the concentration of one component in an equilibrium rxn is increased or decreased.

CO (aq) + 2H2 (aq) ↔ CH4 (aq) + H20 (l)

For example, how would we shift the rxn if we IncreaseThe concentration of Carbon Monoxide in this rxn?

Again we’ll look to Le Chatelier for answers …

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Effect of ConcentrationEffect of ConcentrationLe Chatrelier’s principle says that if we increase the COConcentration the system will respond by trying to Counteract the effect (i.e reduce the CO concentration).

So in our example if we inject more CO into the system Le Chatelier predicts the rxn would shift to the right

CO (aq) + 2H2 (aq) ↔ CH4 (aq) + H20 (l)

NOTE! If we inc [CO] then [H2] dec, [CH4] inc & [H20] inc

Can you explain why [H2] dec?

CO is consumedRf > Rr

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Effect of ConcentrationEffect of ConcentrationIf we decrease the conc. of CH4 Le Chatelier would predictthe rxn would again shift to the right…

CO (aq) + 2H2 (aq) ↔ CH4 (aq) + H20 (l)

And as the [CH4] dec. [H2] dec. [CO] dec. & [H20] inc.

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Consider the following reaction:

H2CO3(aq) + 2NaOH(aq) +Heat ↔ H2O(l)+CO2(g)+Na2O(s)

What happens as you increase:PressureTemperature[NaOH][H2O]

What happens when you decrease:

[CO2] i.e. P[H2CO3]

A PROBLEMA PROBLEM

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Removal of Reactant/ProductRemoval of Reactant/Product

Double replacement reactions involve ionic compoundsThat ionize in water and exchange ions in the process.

KNO3 (aq) + NaCl (aq) ↔ KCl (aq) + NaNO3 (aq)

This is an equilibrium reaction. At equilibrium ALL ions (products and reactants) are present.

(If I evaporated the water ALL compounds would be left behind)

If however one of the products is no longer availableTo participate in the equilibrium rxn, the reaction willProceed left to right to completion.

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Removal of Reactant/ProductRemoval of Reactant/Product

How can we drive these reactions to completion?

One of three things must occur:

1) A PRECIPITATE is produced (insoluble compound)

2) A GAS is Produced that can leave the system

3) A MOLECULE is produced (Covalent compound) - non-ionizable and thus removed from further rxn - typial example is water (H20)

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Haber ProcessHaber ProcessThe Haber process is named after the German ChemistFritz Haber and was developed in 1909 & commercializedIn 1911 to produce NH3, needed for the production ofexplosives during WWI

N2 + 3H2 ↔ 2NH3 + HeatWhat could be done to Increase the yieldof Ammonia in the Haber Process? Father of Chemical

Warfare and NobelLaureate (1918)

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Inc. the Yield of NHInc. the Yield of NH33

We’ll see next that if we decrease temperature toProduce more ammonia we also Decrease reaction rate.Decreasing temp causes the Haber process to shift tothe right but at a slower rate.

N2 + 3H2 ↔ 2NH3 + Heat