FFR: Le Chatelier’s Principle

The previous sections helped you to determine a lot about how a reaction will proceed as it currently exists. What would happen, then, if you were to change something? Say a reaction was at equilibrium, and then you added some reactant. Would would occur?

To answer that, we turn to Le Chatelier’s Principle. The principle states that: If a system currently at equilibrium undergoes a change in either Concentration, Temperature, Pressure, or Volume, then the system will undergo chemical reactions to counteract this change until equilibrium is re-established.

Basically, when something changes, the equilibrium will no longer be held true, and therefore the chemicals must react to fix this.

Let’s start by looking at concentration. If you have a solution with NaCl and AgNO3, and they are in equilibrium, reacting by the following reaction:

$NaCl_{(aq)}+AgNO_{3(aq)}\leftrightarrow AgCl(s)+Na^+_{(aq)}+NO^-_{3(aq)}$

So as to not get too confusing, let’s remove the spectator ions:

$Ag^+_{(aq)}+Cl^-_{(aq)}\leftrightarrow AgCl_{(s)}$

Now say you add some more Chlorine into the solution. That pushes the equilibrium out of whack, as you now have too many reactants. Therefore, more AgCl must form in order to reach equilibrium once more. Once that happens, the reactants are lowered, resulting in equilibrium being reached once more.

Now for the more complicated changes that would result in re-establishment of equilibrium. Let’s start off with temperature.

Temperature having an effect on equilibrium is a bit confusing as we have yet to see the impact of temperature. In fact, the Equilibrium Constant of a reaction relies on temperature, and different temperatures result in different Equilibrium Constants.

In order to predict what happens when temperature changes, we need to look at a new aspect of the reaction, it’s enthalpy change ( $\Delta H$ ). You may recall from Chemistry I that enthalpy is a measurement of heat, and the enthalpy change of a reaction is the heat taken up or released by the reaction. If you have a $\Delta H > 0$, then your reaction is endothermic, and energy has to be added for the reaction to progress. Conversely, a reaction with a $\Delta H < 0$ is exothermic, meaning it releases energy upon progressing.

Therefore, if you know whether a reaction is endothermic or exothermic, you can add an energy term to the equation. For example, a solid melting is an endothermic reaction. Therefore, it can be written like this:

$A_{(s)}+Energy\leftrightarrow A_{(l)}$

Now it is easy to tell that if this reaction is initially at equilibrium, and you add energy, then more of A will have to melt for equilibrium to be re-established.

Instead, let’s look at a different reaction. A + B > C + D has a $\Delta H = -121\frac{kJ}{mol}$. This means that it is exothermic, so it releases energy. This means energy is a product. Therefore, it we were to increase the energy by raising the temperature, then equilibrium will shift to the left, creating more of A and B.

That covers Temperature, so let’s move on to Volume. Volume typically only modifies gaseous systems, so let’s look at those.

Say we have a system at a certain volume containing Nitrogen Monoxide and Oxygen gas, which react to produce Nitrogen Dioxide. The reaction is written as:

$2NO_{(g)}+O_{2(g)}\leftrightarrow 2NO_{2(g)}$

The ideal gas law tells us that is we change volume then we have to also change pressure. Therefore the partial pressures of each component will change, so we can determine which way the reaction moves. So which way would this reaction move if we decrease the volume?

Well let’s look at the expression for this reaction’s Equilibrium Constant.

$K_p = \frac{p^2_{NO_2}}{p^2_{NO}p_{O_2}}$

Now let’s figure out what will happen if we decrease the volume to 80% of its original size. Let’s start by figuring out the change in pressure:

$P_1V_1=P_2V_2$

$P_1V_1=P_2(0.8\cdot V_1)$

$\frac{P_1}{0.8}=P_2$

$1.25\cdot P_1=P_2$

With this relationship, let’s determine the Quotient Constant for the new state with the reduced volume.

$Q_p = \frac{( p_{NO_2}\cdot 1.25)^2}{(p_{NO}\cdot 1.25)^2p_{O_2}\cdot 1.25}$

$Q_p = \frac{p^2_{NO_2}}{1.25\cdot p^2_{NO}p_{O_2}}$

$Q_p = \frac{K_p}{1.25}$

This means that the Quotient Constant is less than the Equilibrium constant, causing the reaction to proceed towards the right.

Okay, well that was straightforward, but it wasn’t really that quick or easy. How can you tell what will happen without going through that whole shebang?

Well the answer lies in the molar ratio of the reactants and products. Let’s look at our equation again.

$2NO_{(g)}+O_{2(g)}\leftrightarrow 2NO_{2(g)}$

Notice that we have three moles of gaseous reactants for every two moles of products. This means we have a molar ratio of 3:2. Now let’s look at the Quotient Constant we determined previously.

$Q_p = \frac{( p_{NO_2}\cdot 1.25)^2}{(p_{NO}\cdot 1.25)^2p_{O_2}\cdot 1.25}$

Let’s pull out that coefficient.

$Q_p = \frac{ p_{NO_2}^2\cdot 1.25^2}{p^2_{NO}p_{O_2}\cdot 1.25^3}$

Notice that the exponents are in the same ratio as the moles. That will always be true, and it means that the side with the higher number of moles will be the predominating side of the equation for determining the impact of a volumetric change.

The other link to make is that an increase in volume will result in a shift towards the dominant side. In other words, the increased volume makes the reaction proceed to generate pressure in the system, by proceeding in such a way as to generate the most gas. In the example case, the reactants had the most moles of gas, meaning that the reaction used products to make reactants.

So what happens when you have a reaction that is equimolar on each side? A reaction like the formation of HCl gas:

$H_{2(g)}+Cl_{2(g)}\leftrightarrow 2HCl_{(g)}$

Well, in this case, a change in volume would actually result in equilibrium being maintained, as both pressures change correspondingly, resulting in the Quotient Constant after a volume change (in either direction) equaling the Equilibrium Constant.

Chapter Problems

1. Carbon Dioxide can react with Hydrogen to form Methane. The reaction is as follows. Name three ways of pushing the equilibrium to the right.
   $CO_{2(g)}+4H_{2(g)}\leftrightarrow CH_{4(g)}+2H_2O_{(g)}$
2. Reviewed replacement: in a 1.00 L closed ideal-gas vessel at a fixed stated temperature, A(g)⇌B(g), K_c=12.4. Initially [A]=0.100 M and [B]=0. Calculate equilibrium [B].
   Review the authored solution

   Let [B]=x and [A]=0.100−x. 12.4=x/(0.100−x), so x=0.0925373 M. All concentrations are nonnegative and their total remains 0.100 M. The original copper/HCl hydrogen-generation premise is superseded.

3. Reviewed replacement: in a constant-volume irreversible model, the methane partial-pressure disappearance law is −dp/dt=kp², with k=0.0422 (atm·s)⁻¹. How long does p fall from 0.400 atm to 0.200 atm?
   Review the authored solution

   t=(1/0.200−1/0.400)/0.0422=59.2417 s. This is a concentration/pressure threshold in a specified kinetic model, rather than time to equilibrium.

4. Ammonia is fixed from atmospheric nitrogen by reacting it with hydrogen. Given that the equilibrium constant is 2.7, if a greenhouse starts out at 1atm, with 70% Nitrogen and 0.5% Hydrogen, what will the partial pressure of ammonia be at equilibrium?
5. Ammonia gas reacts with oxygen according to the following unbalanced equation. A reactor held at 10L and 25ºC contains 4mol of Ammonia and 5mol of Oxygen gas. What is the total pressure of the container at equilibrium?
   $NH_{3(g)}+O_{2(g)}\leftrightarrow NO_{(g)}+H_2O_{(g)}$ $K = 2.22$
6. Reviewed replacement: an irreversible first-order disappearance model for a gas has k=32.0 min⁻¹. How long does its partial pressure fall from 0.100 atm to 0.0100 atm at fixed volume and temperature?
   Review the authored solution

   t=ln(0.100/0.0100)/32.0=0.0719558 min=4.31735 s. No equilibrium-time or tanker-purity prediction is inferred.

7. Hydrogen Cyanide gas can be produced by reacting Methane and Ammonia. The reaction has a $\Delta H=251 \frac {kJ}{mol}$.
   $CH_{4(g)}+NH_{3(g)}\leftrightarrow HCN_{(g)}+3H_{2(g)}$
   The system is currently at equilibrium. Declare what happens under the following conditions:
   a) System volume increases
   b) Temperature increases
   c) Pressure of Ammonia increases
8. Reviewed replacement: A(aq)⇌B(aq) has K_c=7.231. An ideal 0.800 L solution initially has [A]=0.600 M and [B]=0. Calculate the moles of B at equilibrium.
   Review the authored solution

   x/(0.600−x)=7.231 gives x=0.527105 M. n(B)=0.800x=0.421684 mol. This complete aqueous model avoids the original missing gas-space volume and mixed-phase convention.

9. Elemental sulfur can be obtained by treating Hydrogen Sulfide gas with Sulfur Dioxide. If a 100L reactor chamber at 20ºC reacts with 10atm Hydrogen Sulfide and 7atm Sulfur Dioxide, how much Sulfur is created at equilibrium?
   $2H_2S_{(g)}+SO_{2(g)}\leftrightarrow 3S_{(s)}+2H_2O_{(g)}$ $K = 0.843$
10. In 1890, a way to purify Nickel was discovered. It involved passing heated Carbon Monoxide over Nickel. The two reacted to form Nickel Tetracarbonyl ($Ni(CO)_4$) gas. at 50ºC, the equilibrium constant is 12.3. A container is initially at 100kPa CO and contains 10g of Nickel. The container is heated to 50ºC, and its volume is 5L. What is the equilibrium mass of Nickel in the container?
11. Carbon Monoxide and Water vapor can react to form Carbon Dioxide and Hydrogen. The reaction has a $\Delta H=-41\frac{kJ}{mol}$, and an equilibrium constant $K= 0.221$. A 1L canister at 5atm and 25ºC contains equal amounts by volume of Carbon Monoxide and Water. What is the equilibrium mass of Hydrogen gas in the canister?
    $CO_{(g)}+H_2O_{(g)}\leftrightarrow CO_{2(g)}+H_{2(g)}$
13. Reviewed replacement: in an irreversible first-order disappearance model, [NO₂⁻] starts at 4.00 M and k=42.1 day⁻¹. Find its half-life. This formal model does not claim to describe the original reversible gas-producing reaction.
    Review the authored solution

    t½=ln 2/42.1=0.0164643 days=23.7086 min. Equilibrium composition and approach time require separate thermodynamic and reversible kinetic data.

14. Reviewed replacement: A(g)+B(g)⇌2 C(g) in an ideal 5.00 L vessel at 273.15 K has K_c=13.21. Initially there is 1.00 mol each A and B and no C. Calculate equilibrium moles and total pressure, using R=0.082057 L atm mol⁻¹ K⁻¹.
    Review the authored solution

    Let the forward extent be ξ mol. √13.21=2ξ/(1−ξ), so ξ=0.645047 mol. Thus n(A)=n(B)=1.00−ξ=0.354953 mol≈0.355 mol each, and n(C)=2ξ=1.290094 mol≈1.29 mol. The total gas amount remains 2.00 mol because Δn=0. P=nRT/V=8.96555 atm≈8.97 atm (three significant figures). This is a complete hypothetical gas-equilibrium task rather than assumed sulfuric-acid vapor chemistry.

15. Reviewed replacement: a specified reaction has ΔH=−481.1 kJ per mole of reaction as written. Independent equilibrium analysis establishes that its forward extent is 0.250 mol. What is the system’s enthalpy change?
    Review the authored solution

    ΔH_system=(0.250)(−481.1)=−120.275 kJ≈−120 kJ at three significant figures. It releases about 120 kJ under constant-pressure conditions with only pressure–volume work. The reaction extent is supplied rather than inferred from the original unbalanced equation.

16. Reviewed replacement: a formal irreversible disappearance law −d[A]/dt=k[A]² has k=4.20 L mol⁻¹ h⁻¹ and [A]₀=0.100 M. How long until [A]=0.0500 M?
    Review the authored solution

    t=(1/0.0500−1/0.100)/4.20=2.38095 h. The balanced chemical reaction and reverse kinetics of the original lead-oxide task are not inferred.

17. Reviewed replacement: 0.0100 mol Zn reacts with excess dilute nonoxidizing acid by Zn(s)+2 H⁺(aq)→Zn²⁺(aq)+H₂(g). Assuming quantitative reaction, how many moles of hydrogen form?
    Review the authored solution

    One mole of Zn forms one mole H₂, so 0.0100 mol H₂ forms. This explicit reviewed metal/nonoxidizing-acid model replaces the original nitric-acid hydrogen premise.

18. Reviewed replacement: an abstract reactant A undergoes irreversible first-order disappearance with k=0.004333 min⁻¹ and [A]₀=0.150 M. Calculate [A] after 10.0 min and 120 min.
    Review the authored solution

    [A]=[A]₀e⁻ᵏᵗ gives 0.143639 M at 10.0 min and 0.0891816 M at 120 min. Heating NaCl solution does not establish the source’s proposed chlorate/hydrogen reaction; this is a stated kinetic model.

19. Reviewed replacement: for a stipulated hot-alkali stoichiometric model, 3 Cl₂+6 OH⁻→ClO₃⁻+5 Cl⁻+3 H₂O. How many moles of Cl₂ react when 99.0% of an initial 0.500 mol OH⁻ is consumed? This is a paper stoichiometry exercise.
    Review the authored solution

    Consumed OH⁻=0.495 mol. The ratio is 3 mol Cl₂ per 6 mol OH⁻, giving 0.2475 mol Cl₂≈0.248 mol. The original unspecified equilibrium and gas-volume conditions are superseded.

20. Reviewed replacement: in an irreversible first-order disappearance model with k=0.3211 day⁻¹, find the time to consume 90.0% of a reactant.
    Review the authored solution

    [A]/[A]₀=0.100, so t=ln(10)/0.3211=7.17093 days. This operational threshold differs from exact equilibrium time.

21. Reviewed replacement: a closed ideal gas system contains H₂(g)+Cl₂(g)⇌2 HCl(g), with K_p=4.19 at the specified temperature. Initial partial pressures are p(H₂)=p(Cl₂)=1.00 atm and p(HCl)=0. Find equilibrium partial pressures in the same convention.
    Review the authored solution

    Let x be the pressure extent. √4.19=2x/(1−x), giving x=0.505801 atm. Thus p(H₂)=p(Cl₂)=0.494199 atm and p(HCl)=1.011601 atm. No nitric-acid hydrogen generation or mixed aqueous/gas calculation is assumed.