Dynamic Chemical Equilibrium Notes

Dynamic Chemical Equilibrium

Definition

Dynamic chemical equilibrium is a state where a reversible reaction's forward and reverse reactions occur at the same rate.
Consequently, the concentrations of reactants and products remain constant.
Represented as: $X + Y \rightleftharpoons A + B$
A reversible reaction is one in which products can be converted back into reactants.

Equilibrium Characteristics

Equilibrium relates to the comparison of products and reactants in a reversible reaction.
Reactions continue in both directions until equilibrium is reached.
Equilibrium is NOT the cessation of reactions; it means the forward and backward reactions occur at the same rate.
At equilibrium, the amounts of reactants and products on either side remain constant.

Rate of Reaction vs. Chemical Equilibrium

Rate of reaction is the speed at which a reaction happens (i.e., time for a set amount of product to be produced).
Rate of reaction and chemical equilibrium aren't directly related, but at equilibrium, the rates of the forward and reverse reactions are equal.

Le Châtelier's Principle

'When an external stress (change in pressure, temperature, or concentration) is applied to a system in dynamic chemical equilibrium, the equilibrium point will change in such a way as to counteract the stress.'

Amount vs. Time Graphs

Reactants (R) convert to Products (P): $A + B \rightleftharpoons C$
Graphs depict the change in the amount of reactants and products over time.
Equilibrium Position:
- Reactants and products are equal (rare occurrence).
- More products than reactants (equilibrium to the right).
- More reactants than products (equilibrium to the left).

Seesaw Analogy

Reactants (R) and Products (P) on a seesaw.
Initial state: More reactants than products.
Disturbance: Adding more reactants.
New equilibrium: Established after the disturbance, with a slightly different balance of reactants and products.

Effect of Adding Water

Reaction: $[CoCl4]^{2-} + 6H2O \rightleftharpoons [Co(H2O)6]^{2+} + 4Cl^{-}$
- Blue
- Pink
Adding water favors the forward reaction, producing pink $[Co(H2O)6]^{2+}$ ions.
By Le Châtelier's Principle (LCP), the system uses up the excess water molecules.

Effect of Adding Concentrated $HCl$

Adding $HCl$ (containing Cl^{-}$ ions) favors the reverse reaction, producing blue [CoCl_4]^{2-} $ions.</li><li>By LCP, the system tries to use up the chloride ions.</li></ul><h5 collapsed="false" seolevelmigrated="true">General Rules</h5><ul><li>When the forward reaction is favored, reactants become products.</li><li>When the reverse reaction is favored, products become reactants.</li></ul><h5 collapsed="false" seolevelmigrated="true">Effect of Temperature Changes</h5><ul><li>Exothermic reaction:$ \Delta H < 0 $</li><li>Endothermic reaction:$ \Delta H > 0 $</li><li>Increase in temperature favors the endothermic reaction.</li><li>Decrease in temperature favors the exothermic reaction.</li></ul><h5 collapsed="false" seolevelmigrated="true">Effect of Change in Concentration</h5><ul><li>Example:$ 2SO2(g) + O2(g) \rightleftharpoons 2SO_3(g) $</li><li>At time$ t_1 $, the system is in equilibrium.</li><li>At time$ t2 $, the concentration of$ SO2 $is increased.</li><li>The forward reaction speeds up, producing more$ SO_3 $.</li><li>The concentrations of$ O2 $and$ SO2 $decrease until equilibrium is re-established.</li><li>The change happens according to mole ratio.</li></ul><h5 collapsed="false" seolevelmigrated="true">Key Considerations</h5><ul><li>Phases (solid, liquid, gas, aqueous).</li><li>Endothermic vs. exothermic.</li><li>Mole ratio.</li><li>Reversible arrows.</li></ul><h5 collapsed="false" seolevelmigrated="true">Effect of a Catalyst on Equilibrium</h5><ul><li>A catalyst increases the rate of both forward and reverse reactions.</li><li>No equilibrium shift occurs.</li><li>A higher rate is achieved, but the equilibrium position does not change.</li><li>Equilibrium is reached quicker when a catalyst is added.</li></ul><h5 collapsed="false" seolevelmigrated="true">Equilibrium Constant (Kc)</h5><ul><li>The equilibrium constant,$ K_c $, is a ratio between the concentration of the products and the reactants.</li><li>It tells the scientist about the equilibrium position of the reaction.</li><li>$ K_c = \frac{[Products]}{[Reactants]} $</li><li>If$ K_c $is very small, the denominator is very large => equilibrium lies to the left.</li><li>If$ K_c $is almost equal to 1 => equilibrium lies to the left.</li><li>If$ K_c > 1 $, there are more products than reactants at equilibrium => equilibrium lies to the right.</li></ul><h5 collapsed="false" seolevelmigrated="true">Calculating$ K_c $</h5><ul><li>Example 1:$ 2CO(g) + O2(g) \rightleftharpoons 2CO2(g) $<ul><li>$ [O_2] = 2 \times 10^{-3} mol \cdot dm^{-3} $</li><li>$ [CO_2] = 4 \times 10^{-3} mol \cdot dm^{-3} $</li><li>$ [CO] = 2 \times 10^{-2} mol \cdot dm^{-3} $</li><li>$ Kc = \frac{[CO2]^2}{[CO]^2[O_2]} = \frac{(4 \times 10^{-3})^2}{(2 \times 10^{-2})^2 (2 \times 10^{-3})} = 2 $</li></ul></li><li>Example 2:$ SO2(g) + NO2(g) \rightleftharpoons SO_3(g) + NO(g) $<ul><li>$ [SO_2] = 0.4 mol \cdot dm^{-3} $</li><li>$ [NO_2] = 0.05 mol \cdot dm^{-3} $</li><li>$ [SO_3] = 3 mol \cdot dm^{-3} $</li><li>$ [NO] = 0.2 mol \cdot dm^{-3} $</li><li>$ Kc = \frac{[SO3][NO]}{[SO2][NO2]} = \frac{(3)(0.2)}{(0.4)(0.05)} = 30 $</li></ul></li></ul><h5 collapsed="false" seolevelmigrated="true">Components of Reactions</h5><ul><li>Components can be solids (s), liquids (l), gases (g), or in solution (aq).</li><li>Solids and pure liquids have constant concentrations and are not included in$ K_c calculations.
Only (aq) & (g) are put into the K_c $expression.</li></ul><h5 collapsed="false" seolevelmigrated="true">RICE Tables</h5><ul><li>RICE (Reaction, Initial, Change, Equilibrium) tables are used to calculate the concentrations of reactants and products at equilibrium.</li></ul><h6 collapsed="false" seolevelmigrated="true">Steps for Using RICE Tables</h6><ol><li>Write down the mole ratio.</li><li>Write down given 'initial' moles.</li><li>Insert given equilibrium moles.</li><li>Use the column with the most information to work out the change in moles.</li><li>Moving across the 'change in moles' row, use the mole ratio to calculate the change in moles of other reagents.</li><li>Work out equilibrium moles for other reagents.</li><li>Calculate concentrations.</li></ol><h6 collapsed="false" seolevelmigrated="true">Example</h6><ul><li> </li><li>Reaction:$ 2NH3(g) \rightleftharpoons N2(g) + 3H_2(g) $</li><li> </li><li>One mole of$ NH3 $is placed in a 2$ dm^3 $container. The equilibrium mixture is analyzed and found to contain 0.6 mol of$ H2 $.</li><li> </li><li>Calculate$ K_c $. <table style="min-width: 125px"><colgroup><col style="min-width: 25px"><col style="min-width: 25px"><col style="min-width: 25px"><col style="min-width: 25px"><col style="min-width: 25px"></colgroup><tbody><tr><th colspan="1" rowspan="1"></th><th colspan="1" rowspan="1">$ 2NH_3 $</th><th colspan="1" rowspan="1">$ N_2 $</th><th colspan="1" rowspan="1">$ 3H_2 $</th><th colspan="1" rowspan="1"> </th></tr><tr><td colspan="1" rowspan="1"></td><td colspan="1" rowspan="1"></td><td colspan="1" rowspan="1"></td><td colspan="1" rowspan="1"></td><td colspan="1" rowspan="1"> </td></tr><tr><td colspan="1" rowspan="1">Initial</td><td colspan="1" rowspan="1">1</td><td colspan="1" rowspan="1">0</td><td colspan="1" rowspan="1">0</td><td colspan="1" rowspan="1"> </td></tr><tr><td colspan="1" rowspan="1">Change</td><td colspan="1" rowspan="1">-0.4</td><td colspan="1" rowspan="1">+0.2</td><td colspan="1" rowspan="1">+0.6</td><td colspan="1" rowspan="1"> </td></tr><tr><td colspan="1" rowspan="1">Equilibrium</td><td colspan="1" rowspan="1">0.6</td><td colspan="1" rowspan="1">0.2</td><td colspan="1" rowspan="1">0.6</td><td colspan="1" rowspan="1"> </td></tr><tr><td colspan="1" rowspan="1">Concentration</td><td colspan="1" rowspan="1">0.3</td><td colspan="1" rowspan="1">0.1</td><td colspan="1" rowspan="1">0.3</td><td colspan="1" rowspan="1"></td></tr><tr><td colspan="1" rowspan="1"><ul><li>$ Kc = \frac{[N2][H2]^3}{[NH3]^2} = \frac{(0.1)(0.3)^3}{(0.3)^2} = 0.03 $</li></ul></td><td colspan="1" rowspan="1"></td><td colspan="1" rowspan="1"></td><td colspan="1" rowspan="1"></td><td colspan="1" rowspan="1"></td></tr><tr><td colspan="1" rowspan="1"></td><td colspan="1" rowspan="1"></td><td colspan="1" rowspan="1"></td><td colspan="1" rowspan="1"></td><td colspan="1" rowspan="1"></td></tr><tr><td colspan="1" rowspan="1"><h5 collapsed="false" seolevelmigrated="true">Industrial Applications</h5></td><td colspan="1" rowspan="1"></td><td colspan="1" rowspan="1"></td><td colspan="1" rowspan="1"></td><td colspan="1" rowspan="1"></td></tr></tbody></table><h6 collapsed="false" seolevelmigrated="true">Haber Process</h6><ul><li>Used to make ammonia for fertilizer and bombs.</li><li>$ N2(g) + 3H2(g) \rightleftharpoons 2NH_3(g) + E \quad \Delta H = -46kJ \cdot mol^{-1} $</li><li>$ K_c = 0.3 $at a temperature of 450°C.</li><li>Conditions:<ul><li>Catalyst: iron oxide$ (Fe2O3) $</li><li>Relatively low temperature: 450°C (favors the exothermic reaction).</li><li>High pressure: 250 atm (favors the least moles).</li></ul></li><li>Assess the effect of temperature and pressure on this equilibrium and comment on the use of a catalyst for yield</li></ul><h6 collapsed="false" seolevelmigrated="true">Ostwald Process</h6><ul><li>Makes nitric acid for fertilizer and labs.</li><li>$ 2NO(g) + O2(g) \rightleftharpoons 2NO2(g) + E \quad \Delta H = -117kJ \cdot mol^{-1} $</li><li>Conditions:<ul><li>Catalyst: platinum.</li><li>Done at low temperatures</li><li>$ NO $gas is cooled before the reaction takes place.</li></ul></li></ul><h6 collapsed="false" seolevelmigrated="true">Contact Process</h6><ul><li>Makes sulfuric acid.</li><li>$ 2SO2(g) + O2(g) \rightleftharpoons 2SO_3(g) + E \quad \Delta H = -95kJ \cdot mol^{-1} $</li><li>$ K_c = 10^{10}$$ at

Dynamic Chemical Equilibrium Notes

Dynamic Chemical Equilibrium

Definition

Equilibrium Characteristics

Rate of Reaction vs. Chemical Equilibrium

Le Châtelier's Principle

Amount vs. Time Graphs

Seesaw Analogy

Effect of Adding Water

Effect of Adding Concentrated HClHClHCl

Effect of Adding Concentrated $HCl$