Tuesday, September 14, 2010

14.4-14.9: Qualitatively Interpreting the Equilibrium Constant-Effect of a Catalyst

Qualitatively Interpreting the Equilibrium Constant:

-If Kc is a large number (example: 4.1x108) the products are favored at equilibrium, and there will be more product than reactant. To obtain Kc, the reactants are in the denominator, and when you divide by a smaller and smaller number (as the reactants dwindle and the products emerge), the value of Kc will get bigger and bigger.

-If Kc is a very small number (example: 4.6x10-31) the reactants are favored at equilibrium, and there will be more reactant than product. Again, to obtain Kc, the products are in the numerator, and when you divide a small number (products) by a larger number (reactants), the value of Kc will be very small.

-If Kc is within an order of magnitude near 1 (example: 7.2), there is an even distribution of reactants and products.

-If Kc = 0, there has been no reaction.

Predicting the Direction of Reaction:

Qc = reaction quotient
-Obtained the same exact way used to obtain Kc. Plug in the concentrations of the products raised to the power of their coefficients, divided by the concentrations of the reactants raised to the power of their coefficients.



-The only difference is Kc is at equilibrium and Qc is not. So we can use Qc to see how to get to Kc.

-If Qc > Kc, numerator is too big, need more reactant, reaction will shift to the left
-If Qc < Kc, denominator is too big, need more product, reaction will shift to the right -If Qc = Kc, already at equilibrium

Calculating Equilibrium Concentrations:

1. Set up a table of concentrations (Initial, Change, and Equilibrium expressions in x)
2. Substitute expressions in x for equilibrium concentrations into the equilibrium-constant equation.
3. Solve the equilibrium-constant equation for the values of the equilibrium concentrations.

Note: If you cannot get rid of a squared x term easily by simply taking the square root of everything (which will happen if your initial concentrations are the same), you will have to use the quadratic equation.

This will yield two answers, only one answer will be correct. If one is positive and the other is negative, the positive answer is the correct answer. If both are positive, plug the answers into your x expressions to see which one is realistically plausible (doesn't yield a negative number for any concentrations).

Changing Reaction Conditions:

-Le Chatelier's Principle: when a system in chemical equilibrium is disturbed by a change in temperature, pressure, or a concentration, the system shifts in equilibrium composition in a way that tends to counteract this change variable.

-Three ways to alter equilibrium concentration of a gaseous reaction mixture:
1. Changing concentrations
2. Changing the partial pressures of the products and reactants by changing the volume
3. Changing the temperature

-Concentration:
-When more reactant is added or product removed from the reaction vessel, the reaction will shift to the right to compensate for this gain/loss.
-When more product is added or reactant removed, the reaction will shift to the left to compensate. More reactants will be produced and amount of product will be subsequently decreased.

-Pressures/Volume:
-Pressure is directly proportional to moles of gas, so a decrease in the volume of a reaction vessel will cause the reaction to move in the direction that has a smaller number of moles of gas.
-An increase in volume will cause the reaction to move in the direction that has a larger number of moles of gas.
-When there are equal numbers of moles of gas on either side of the reaction (example: 2 total moles of reactant yields 2 total moles of product), then a change in pressure causes no shift in the reaction.

-Temperature:
-Must know the change in enthalpy of a reaction (ΔH) to know how temperature will shift the reaction.
-If ΔH is negative, the reaction is exothermic. Heat can be "imagined" to be a product of the reaction, and when you increase a product, the reaction will shift to the left to compensate. So adding heat to an exothermic reaction will cause it to shift to the left. And a temperature decrease will cause the reaction to shift to the right.
-If ΔH is positive, the reaction is endothermic. Heat can now be imagined to be a reactant. When you increase a reactant, the reaction will shift to the right to compensate. So adding heat to an endothermic reaction will cause it to shift to the right. And a temperature decrease will cause the reaction to shift to the left.

Effect of a Catalyst:

Kc is unaffected by a catalyst. The rate of reaction increases, but the final composition at equilibrium will be the same!

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