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: $[CoCl<em>4]^{2-} + 6H</em>2O \rightleftharpoons [Co(H<em>2O)</em>6]^{2+} + 4Cl^{-}$

  • Blue

  • Pink

• Adding water favors the forward reaction, producing pink $[Co(H<em>2O)</em>6]^{2+}$ 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-}$ions.</p></li><li><p>By LCP, the system tries to use up the chloride ions.</p></li></ul><h5 collapsed="false" seolevelmigrated="true">General Rules</h5><ul><li><p>When the forward reaction is favored, reactants become products.</p></li><li><p>When the reverse reaction is favored, products become reactants.</p></li></ul><h5 collapsed="false" seolevelmigrated="true">Effect of Temperature Changes</h5><ul><li><p>Exothermic reaction:$\Delta H < 0$</p></li><li><p>Endothermic reaction:$\Delta H > 0$</p></li><li><p>Increase in temperature favors the endothermic reaction.</p></li><li><p>Decrease in temperature favors the exothermic reaction.</p></li></ul><h5 collapsed="false" seolevelmigrated="true">Effect of Change in Concentration</h5><ul><li><p>Example:$2SO2(g) + O2(g) \rightleftharpoons 2SO_3(g)$</p></li><li><p>At time$t_1$, the system is in equilibrium.</p></li><li><p>At time$t2$, the concentration of$SO2$is increased.</p></li><li><p>The forward reaction speeds up, producing more$SO_3$.</p></li><li><p>The concentrations of$O2$and$SO2$decrease until equilibrium is re-established.</p></li><li><p>The change happens according to mole ratio.</p></li></ul><h5 collapsed="false" seolevelmigrated="true">Key Considerations</h5><ul><li><p>Phases (solid, liquid, gas, aqueous).</p></li><li><p>Endothermic vs. exothermic.</p></li><li><p>Mole ratio.</p></li><li><p>Reversible arrows.</p></li></ul><h5 collapsed="false" seolevelmigrated="true">Effect of a Catalyst on Equilibrium</h5><ul><li><p>A catalyst increases the rate of both forward and reverse reactions.</p></li><li><p>No equilibrium shift occurs.</p></li><li><p>A higher rate is achieved, but the equilibrium position does not change.</p></li><li><p>Equilibrium is reached quicker when a catalyst is added.</p></li></ul><h5 collapsed="false" seolevelmigrated="true">Equilibrium Constant (Kc)</h5><ul><li><p>The equilibrium constant,$K_c$, is a ratio between the concentration of the products and the reactants.</p></li><li><p>It tells the scientist about the equilibrium position of the reaction.</p></li><li><p>$K_c = \frac{[Products]}{[Reactants]}$</p></li><li><p>If$K_c is very small, the denominator is very large => equilibrium lies to the left.

• If K_c is almost equal to 1 => equilibrium lies to the left.

• If K_c > 1, there are more products than reactants at equilibrium => equilibrium lies to the right.

Calculating K_c$</h5><ul><li><p>Example 1:$2CO(g) + O2(g) \rightleftharpoons 2CO2(g)$</p><ul><li><p>$[O_2] = 2 \times 10^{-3} mol \cdot dm^{-3}$</p></li><li><p>$[CO_2] = 4 \times 10^{-3} mol \cdot dm^{-3}$</p></li><li><p>$[CO] = 2 \times 10^{-2} mol \cdot dm^{-3}$</p></li><li><p>$Kc = \frac{[CO2]^2}{[CO]^2[O_2]} = \frac{(4 \times 10^{-3})^2}{(2 \times 10^{-2})^2 (2 \times 10^{-3})} = 2$</p></li></ul></li><li><p>Example 2:$SO2(g) + NO2(g) \rightleftharpoons SO_3(g) + NO(g)$</p><ul><li><p>$[SO_2] = 0.4 mol \cdot dm^{-3}$</p></li><li><p>$[NO_2] = 0.05 mol \cdot dm^{-3}$</p></li><li><p>$[SO_3] = 3 mol \cdot dm^{-3}$</p></li><li><p>$[NO] = 0.2 mol \cdot dm^{-3}$</p></li><li><p>$Kc = \frac{[SO3][NO]}{[SO2][NO2]} = \frac{(3)(0.2)}{(0.4)(0.05)} = 30$</p></li></ul></li></ul><h5 collapsed="false" seolevelmigrated="true">Components of Reactions</h5><ul><li><p>Components can be solids (s), liquids (l), gases (g), or in solution (aq).</p></li><li><p>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.</p></li></ul><h5 collapsed="false" seolevelmigrated="true">RICE Tables</h5><ul><li><p>RICE (Reaction, Initial, Change, Equilibrium) tables are used to calculate the concentrations of reactants and products at equilibrium.</p></li></ul><h6 collapsed="false" seolevelmigrated="true">Steps for Using RICE Tables</h6><ol><li><p>Write down the mole ratio.</p></li><li><p>Write down given 'initial' moles.</p></li><li><p>Insert given equilibrium moles.</p></li><li><p>Use the column with the most information to work out the change in moles.</p></li><li><p>Moving across the 'change in moles' row, use the mole ratio to calculate the change in moles of other reagents.</p></li><li><p>Work out equilibrium moles for other reagents.</p></li><li><p>Calculate concentrations.</p></li></ol><h6 collapsed="false" seolevelmigrated="true">Example</h6><p></p><ul><li><p><br></p></li><li><p>Reaction:$2NH3(g) \rightleftharpoons N2(g) + 3H_2(g)$</p></li><li><p><br></p></li><li><p>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$H2$.</p></li><li><p><br></p></li><li><p>Calculate$K_c$.<br></p><p></p><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"><p></p></th><th colspan="1" rowspan="1"><p>$2NH_3$</p></th><th colspan="1" rowspan="1"><p>$N_2$</p></th><th colspan="1" rowspan="1"><p>$3H_2$</p></th><th colspan="1" rowspan="1"><p><br></p></th></tr><tr><td colspan="1" rowspan="1"><p></p></td><td colspan="1" rowspan="1"><p></p></td><td colspan="1" rowspan="1"><p></p></td><td colspan="1" rowspan="1"><p></p></td><td colspan="1" rowspan="1"><p><br></p></td></tr><tr><td colspan="1" rowspan="1"><p>Initial</p></td><td colspan="1" rowspan="1"><p>1</p></td><td colspan="1" rowspan="1"><p>0</p></td><td colspan="1" rowspan="1"><p>0</p></td><td colspan="1" rowspan="1"><p><br></p></td></tr><tr><td colspan="1" rowspan="1"><p>Change</p></td><td colspan="1" rowspan="1"><p>-0.4</p></td><td colspan="1" rowspan="1"><p>+0.2</p></td><td colspan="1" rowspan="1"><p>+0.6</p></td><td colspan="1" rowspan="1"><p><br></p></td></tr><tr><td colspan="1" rowspan="1"><p>Equilibrium</p></td><td colspan="1" rowspan="1"><p>0.6</p></td><td colspan="1" rowspan="1"><p>0.2</p></td><td colspan="1" rowspan="1"><p>0.6</p></td><td colspan="1" rowspan="1"><p><br></p></td></tr><tr><td colspan="1" rowspan="1"><p>Concentration</p></td><td colspan="1" rowspan="1"><p>0.3</p></td><td colspan="1" rowspan="1"><p>0.1</p></td><td colspan="1" rowspan="1"><p>0.3</p></td><td colspan="1" rowspan="1"><p></p></td></tr><tr><td colspan="1" rowspan="1"><ul><li><p>$Kc = \frac{[N2][H2]^3}{[NH3]^2} = \frac{(0.1)(0.3)^3}{(0.3)^2} = 0.03$</p></li></ul></td><td colspan="1" rowspan="1"><p></p></td><td colspan="1" rowspan="1"><p></p></td><td colspan="1" rowspan="1"><p></p></td><td colspan="1" rowspan="1"><p></p></td></tr><tr><td colspan="1" rowspan="1"><p></p></td><td colspan="1" rowspan="1"><p></p></td><td colspan="1" rowspan="1"><p></p></td><td colspan="1" rowspan="1"><p></p></td><td colspan="1" rowspan="1"><p></p></td></tr><tr><td colspan="1" rowspan="1"><h5 collapsed="false" seolevelmigrated="true">Industrial Applications</h5></td><td colspan="1" rowspan="1"><p></p></td><td colspan="1" rowspan="1"><p></p></td><td colspan="1" rowspan="1"><p></p></td><td colspan="1" rowspan="1"><p></p></td></tr></tbody></table><h6 collapsed="false" seolevelmigrated="true">Haber Process</h6><ul><li><p>Used to make ammonia for fertilizer and bombs.</p></li><li><p>$N2(g) + 3H2(g) \rightleftharpoons 2NH_3(g) + E \quad \Delta H = -46kJ \cdot mol^{-1}$</p></li><li><p>$K_c = 0.3$at a temperature of 450°C.</p></li><li><p>Conditions:</p><ul><li><p>Catalyst: iron oxide$(Fe2O3)$</p></li><li><p>Relatively low temperature: 450°C (favors the exothermic reaction).</p></li><li><p>High pressure: 250 atm (favors the least moles).</p></li></ul></li><li><p>Assess the effect of temperature and pressure on this equilibrium and comment on the use of a catalyst for yield</p></li></ul><h6 collapsed="false" seolevelmigrated="true">Ostwald Process</h6><ul><li><p>Makes nitric acid for fertilizer and labs.</p></li><li><p>$2NO(g) + O2(g) \rightleftharpoons 2NO2(g) + E \quad \Delta H = -117kJ \cdot mol^{-1}$</p></li><li><p>Conditions:</p><ul><li><p>Catalyst: platinum.</p></li><li><p>Done at low temperatures</p></li><li><p>$NO$gas is cooled before the reaction takes place.</p></li></ul></li></ul><h6 collapsed="false" seolevelmigrated="true">Contact Process</h6><ul><li><p>Makes sulfuric acid.</p></li><li><p>$2SO2(g) + O2(g) \rightleftharpoons 2SO_3(g) + E \quad \Delta H = -95kJ \cdot mol^{-1}$</p></li><li><p>$K_c = 10^{10}$$ at