WebLecture

- Systems of Reactions
- Equilibrium
- Practice with Concepts
- Discussion Questions
- Optional Website Reading

In earlier discussions, we've seen how we can break a chemical reaction down into individual steps. The sum of reactants, products, and the enthalpy of reaction for the overall or net reaction is the sum of the individual reactions. We could reverse an equation to get a desired net reaction.

Suppose that we have done experiments that determine the enthalpy of the following reactions:

Reaction #1: | C(s) + O_{2}(g) → CO_{2}(g) | ΔH_{f} = -393.5 kJ/mole |

Reaction #2: | S(s) + O_{2}(g) → SO_{2}(g) | ΔH_{f} = -296.8 kJ/mole |

Reaction #3: | C(s) + 2 S(s) → CS_{2}(l) | ΔH_{f} = 87.9 kJ/mole |

We want to have the net overall reaction:

Reaction #4: CS_{2}(l) + 3 O_{2}(g) → CO_{2}(g) + 2 SO_{2}(g)

In order to achieve this, we need a reaction with 2 SO_{2} in the products, so we have to double reaction #2, which also doubles its enthalpy production.

We also need a reaction with CS_{2} in the reactants, so we need to reverse reaction #3, which makes its reactants products, but also turns the endothermic reaction into an exothermic reaction.

C(s) + O_{2}(g) → CO_{2}(g) | ΔH_{f} = -393.5 kJ/mole |

2 S(s) + 2 O_{2}(g) → 2 SO_{2}(g) | ΔH_{f} = (2)(-296.8 kJ/mole) |

CS_{2}(l) → C(s) + 2 S(s) | ΔH_{f} = -87.9 kJ/mole |

Total Reaction: | |

C + 2S + 3O_{2} + CS_{2} → CO_{2} + 2 SO_{2} + C + 2S | |

Net Reaction: | |

3O_{2} + CS_{2} → CO_{2} + 2 SO_{2} | -393.5+ (2)(-296.8) +(-87.9) = -1076.8 J/mole |

Now we do the same thing, creating systems of reactions, but this time, we want to see what happens ot the equilibrium constant. We need to consider what happens to the equilibrium constant when we

- multiply an equation by a stoichiometric coefficient
- reverse the equation direction

First, we look at the effect of doubling an equation's coefficients. We know from earlier considerations that an equation like

aA + Bb → cC gives us an equilibrium constant

$$K\text{}=\text{}\frac{{[C]}^{c}}{{[A]}^{a}{[B]}^{b}}$$So our exponents on the concentrations depend on the stoichiometric coefficients of the reaction. If we have the reaction

2H_{2} + O_{2} → 2H_{2}O with the equilibrium constant

$${K}_{1}\text{}=\text{}\frac{{[{H}_{2}O]}^{2}}{{[{H}_{2}]}^{2}{[{O}_{2}]}^{\mathit{1}}}$$ and we double it to

4H_{2} + 2O_{2} → 4H_{2}O the equilibrium exponents change as well to

$${K}_{2}\text{}=\text{}\frac{{[{H}_{2}O]}^{4}}{{[{H}_{2}]}^{4}{[{O}_{2}]}^{2}}\text{}=\text{}{({K}_{1})}^{2}$$

**The equilibrium constant for a reaction multiplied by X is equal to raising the original equilibrium value to the exponent X: K _{2} = (K_{1})^{x}.**

We can do the same kind of analysis for the situations where we we need to reverse an equation. If we start with our initial water composition reaction

2H_{2} + O_{2} → 2H_{2}O with the equilibrium constant

$${K}_{1}\text{}=\text{}\frac{{[{H}_{2}O]}^{2}}{{[{H}_{2}]}^{2}{[{O}_{2}]}^{\mathit{1}}}$$ and we reverse it to

2H_{2}O → 2H_{2} + O_{2}, we've changed the products and reactants, so the new equilibrium constant is

$${K}_{3}\text{}=\text{}\frac{{[{H}_{2}]}^{2}{[{O}_{2}]}^{1}}{{[{H}_{2}O]}^{2}}\text{}=\text{}\frac{1}{{K}_{1}}$$

**The equilibrium constant for a reversed reaction is the inverse of its value for the forward reaction: K _{3} = 1/K_{1}.**

If we add reactions, the effect on the concentrations at equilibrium is equivalent to multiplying the equilibrium values together.

**The equilibrium constant two reactions with equilibrium constants K _{A} and K_{B} is equal to the product of the constants: K_{net} = K_{A}*K_{B}.**

Any system in a given state possesses inertia, resistance to change to its state. Systems in equilibrium in particular will react to an impulse to change by resisting the change to reduce its effects.

For example, if we have a chemical reaction

aA + bB → cC

at equilibirum, so that its concentrations are no longer changing, and we add product C to the system, the reaction will run *backwards* to reduce the increased concentration of C and establish a new equilibrium as close to the original ratio of products to reactants as possible.

The same thing happens if we change volumes for gases in equilibrium. The total number of particles determines the pressure of the gas, and the system will try to maintain this pressure, which is a result of the energy state of the gas. If we have a system where one gas converts to another in the ratio 2A → B, and we cut the volume in half, the reaction will run forward to decrease the number of A molecules until a new total number of molecules is reached that approximates the original pressure as closely as possible.

We also need to take into account the enthalpy of the reaction. If we have an endothermic reaction that takes in heat, and we add heat to the system, the reaction will run forward to reduce the heat content of the system and restore the original temperature as closely as possible. If we cool an endothermic reaction, it will run backwards (if it can) to restore heat to its system.

All of these processes support the fundamental principle that energy must be conserved, and systems will resist changes to their energy state if they can.

Note that the new equilibrium positions won't be exactly the same as the original equilibrium ratios in some cases.

- You can review the basics of Le Chatelier's principle at the Chem Guide site.
- There is a good interative animmation on Le Chatelier's principle at Learners TV

A list of likely websites that can help.

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