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) + O2(g) → CO2(g)||ΔHf = -393.5 kJ/mole|
|Reaction #2:||S(s) + O2(g) → SO2(g)||ΔHf = -296.8 kJ/mole|
|Reaction #3:||C(s) + 2 S(s) → CS2(l)||ΔHf = 87.9 kJ/mole|
We want to have the net overall reaction:
Reaction #4: CS2(l) + 3 O2(g) → CO2(g) + 2 SO2(g)
In order to achieve this, we need a reaction with 2 SO2 in the products, so we have to double reaction #2, which also doubles its enthalpy production.
We also need a reaction with CS2 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) + O2(g) → CO2(g)||ΔHf = -393.5 kJ/mole|
|2 S(s) + 2 O2(g) → 2 SO2(g)||ΔHf = (2)(-296.8 kJ/mole)|
|CS2(l) → C(s) + 2 S(s)||ΔHf = -87.9 kJ/mole|
|C + 2S + 3O2 + CS2 → CO2 + 2 SO2 + C + 2S|
|3O2 + CS2 → CO2 + 2 SO2||-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
First, we look at the effect of doubling an equation's coefficients. We know from earlier considerations that an equation like
So our exponents on the concentrations depend on the stoichiometric coefficients of the reaction. If we have the reaction
The equilibrium constant for a reaction multiplied by X is equal to raising the original equilibrium value to the exponent X: K2 = (K1)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
The equilibrium constant for a reversed reaction is the inverse of its value for the forward reaction: K3 = 1/K1.
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 KA and KB is equal to the product of the constants: Knet = KA*KB.
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
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.
A list of likely websites that can help.
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