Chemical Equilibrium PDF

**[Chemical Equilibrium]** \*\*Definition: Chemical equilibrium is a state of dynamic balance in a reversible reaction where the rate of the forward reaction is the same as the rate of the backward reaction.\*\* A reversible reaction is one in which the products react to give back the reactants, i...

**[Chemical Equilibrium]** \*\*Definition: Chemical equilibrium is a state of dynamic balance in a reversible reaction where the rate of the forward reaction is the same as the rate of the backward reaction.\*\* A reversible reaction is one in which the products react to give back the reactants, i.e., the reaction is going in both directions. \*\*Definitions: In a dynamic state the reactants are continuously forming products and the products are continuously forming reactants.\ In a dynamic equilibrium the rate of the forward reaction is equal to the rate of the backward reaction.\*\* **[Le Chatelier's Principle]** In 1888, Henri Le Chatelier, a French Chemist, put forward a rule that allowed chemists to predict the direction taken by an equilibrium reaction if the conditions of the reaction were changed. \*\*Definition: Le Chatelier's Principle states that if a stress is applied to a system at equilibrium, the system re-adjusts to relieve the stress applied.\*\* Stress simply means a change in the conditions of the reaction, e.g., concentration, pressure, temperature. For example: 2NO~2(g)~ N~2~O~4(g)~ ΔH = -58.1 kj mol^-1^\ 2NO~2(g)~ N~2~O~4(g)~ ΔH = +58.1 kj mol^-1^ In the forward reaction, heat is given out so it is an exothermic reaction. In the backward reaction, heat is taken in making it an endothermic reaction.\ ΔH -- means that a reaction has given out heat\ ΔH + means that a reaction has taken in heat **[Effect of Change of Pressure on Reactions Involving Gases]** Le Chatelier's Principle predicts that in an all-gaseous reaction an increase in pressure will favour the reaction that brings about a reduction in volume, i.e., it will shift towards the side with the smaller number of molecules. \- A change in the pressure on a system affects only an equilibrium reaction that has an unequal numbers of gaseous reactants and products.\ - Equilibrium will shift towards the side with the smaller number of molecules.\ - The smaller number of molecules occupies less volume and this helps to absorb the stress of increased pressure \- A catalyst speeds up the rate at which equilibrium is reached but it does not change the position of equilibrium. The following reaction occurs in a closed vessel at a certain temperature. The forward reaction is exothermic\ PCl~3(g)~ + Cl~2(g)~ ↔ PCl~5(g)~ What would be the effect on the concentration of PCl~5~, in the equilibrium mixture of:\ (i) introducing more PCl~3~ into the mixture\ (ii) increasing the pressure\ (iii) increasing the temperature? \(i) If more PCl~3~ is introduced into the container, the equilibrium will shift to the right to absorb the increase. The concentration of PCl~5~ will increase \(ii) All of the substances are gaseous. Two molecules on the left-hand side but only one on the right hand side. If we increase the pressure, the equilibrium shifts to the side that has the smaller number of molecules. The equilibrium shifts to the right-hand side. The concentration of PCl~5~ increases. \(iii) If the forward reaction is exothermic then the backward reaction is endothermic. If the temperature is increases, the equilibrium will shift in the backwards direction in order to absorb this heat, The concentration of PCl~5~ will decrease. **[Mandatory Experiment: To illustrate Le Chatelier's Principle using the reaction between iron (III) chloride and potassium thiocyanate]** 1. Mix together about 5 cm^3^ respectively of solutions of iron(III) chloride and potassium thiocyanate in a beaker. Note the formation of the red complex. Fe^3+^ + CNS^-^ Fe(CNS)^2+^ yellow red Since the red complex above is formed, the equilibrium must lie on the right hand side of the equation. 2. Divide the mixture into three portions in separate boiling tubes. Keep one of these as a control. 3. Using a fume cupboard, add some concentrated hydrochloric acid to the second tube until the red colour disappears. In keeping with Le Chatelier\'s Principle, the red colour disappears as the equilibrium is shifted to the left hand side to replace the Fe^3+^ ions removed. 4. Add an equivalent amount of water to the third tube and compare. **This comparison should indicate that the extent of lightening of the colour is not due to a diluting effect.** 5. To the second tube, add some potassium thiocyanate solution. The red complex reforms because the equilibrium is shifted to the right **[Industrial Applications of Le Chatelier's Principle]** [Manufacture of Ammonia by the Haber Process] \- Ammonia is one of the most important chemicals manufactured by the chemical industry. It is used o manufacture fertilisers, explosives, and cleaning agent Ammonia is manufactured by reacting nitrogen with hydrogen N~2~ + 3H~2(g)~ [\$\\frac{\\text{Fe}}{\\leftrightarrow}\$]{.math.inline} *2NH~3(g)~ ΔH = -92 kj mol^-1^* *- The amount of Ammonia produced depends on the temperature and the pressure inside the reaction vessel.\ - A pressure of approximately 200 atmospheres is used in the production of Ammonia\ - The above reaction is exothermic\ - A temperature of approximately 200^o^C is used in the production of Ammonia* *Le Chatelier's Principle tells us that best conditions to maximise the yield of ammonia in the Haber process are high pressure and low temperature* *[Manufacture of Sulfuric Acid by the Contact Process]* *- Sulfuric Acid is manufactured by a process known as the Contact Process.* *2SO~2(g)~* + O~2(g)~ [\$\\frac{V\_{2}O\_{5}}{\\leftrightarrow}\$]{.math.inline} *2SO~3~ ΔH = -196 kj mol^-1^* *- Sulfuric acid is made by dissolving the sulfur trioxide in the above reaction in water.\ - A pressure just above atmospheric pressure is used to produce Sulfuric Acid\ - A temperature of 450^o^C is the best temperature for the catalyst to work at in the Contact Process* ***[The Equilibrium Constant]*** K~c~ is called the equilibrium constant and represents the relationship between the concentrations of the reactant and products of a system at equilibrium. K~c~ = [\$\\frac{{\\left\\lbrack C \\right\\rbrack\\ }\^{\\text{c\\ \\ }{\\lbrack D\\rbrack}\^{d}}}{{\\lbrack A\\rbrack}\^{a}{\\lbrack B\\rbrack}\^{b}}\$]{.math.inline} K~c~ = K is the constant, c is the concentration\ \[\] = concentration measured in moles per litre \- The products are always written above the line and the reactants are always below the line\ - The concentrations of the products are multiplied together\ - The concentrations of the reactants are multiplied together\ - The concentration of each species is raised to the power of its coefficient in the balanced equation K~c~ = [\$\\frac{{\\lbrack NH}\_{{3\\rbrack}\^{2}}}{{\\lbrack N}\_{2\\rbrack}\\ \\lbrack H\_{2\\rbrack\^{3}}}\$]{.math.inline} The reaction:\ N~2(g)~ + 3H~2(g)~ ↔ 2NH~3(g)\ ~was allowed to reach equilibrium at 450^o^C in a one litre container. It was found that there were 13.6 moles N~2~, 1.0 mole H~2~ and 1.5 moles NH~3~ in the container when equilibrium had been established. Calculate the value of K~c~ at this temperature. K~c~ = [\$\\frac{{\\lbrack NH}\_{{3\\rbrack}\^{2}}}{{\\lbrack N}\_{2\\rbrack}\\ \\lbrack H\_{2\\rbrack\^{3}}}\$]{.math.inline} K~c~ = [\$\\frac{{(1.5)}\^{2}}{{(13.6)(1.0)}\^{3}}\$]{.math.inline} = 0.165 K~c~ = 0.165 **[Exam Questions]** [2014 -- HL -- Section B -- Question 9] Consider the following reversible reaction\ 2NO (g) ֕ N~2~ (g) + O~2~ (g) ∆H = --183 kJ that has an equilibrium constant (Kc) value of 20.25 at a certain high temperature T.\ (a) Write the equilibrium constant expression for the reaction. K~c~ = [\$\\frac{{\\lbrack NH}\_{{3\\rbrack}\^{2}}}{{\\lbrack N}\_{2\\rbrack}\\ \\lbrack H\_{2\\rbrack\^{3}}}\$]{.math.inline}\ (b) Calculate the number of moles of nitrogen gas (N~2~) in the reaction mixture at equilibrium when a 2 mole sample of nitrogen monoxide decomposes to nitrogen gas and oxygen gas in a closed container at temperature T. 2NO N~2~ O~2~ -------------------- --------- ------ ------ Initial 2 0 0 Change -2x +x +x At equilibrium (L) 2 -- 2x x x [\$\\frac{{(x)}\^{2}}{{(2 - 2x)}\^{2}}\$]{.math.inline} = 20.25 [\$\\frac{x}{2 - x}\$]{.math.inline} = 4.5 X = [\$\\sqrt{4.5}\$]{.math.inline} X = 0.9 moles \(c) State Le Châtelier's principle. If system at equilibrium is subjected to a stress it tends to oppose the stress\ What effect, if any, would an increase in\ (i) the temperature, Value decreases - Reaction shifts in an endothermic direction to oppose the added heat\ (ii) the pressure, have on the value of Kc for this reaction? No effect - Only temperature change affects K~c~\ Justify your answer in each case.\ (d) This reaction is one of several that occur in the catalytic converters fitted to car exhausts. Since the exhaust gases are in the catalytic converter of the car for a very short time (0.1 -- 0.4 seconds), the rate of reaction must be very high. Name two of the metals used as catalysts in catalytic converters. Platinum, palladium and rhodium\ What type of catalysis occurs? Heterogenous\ Give one way that the catalysts increase the rate of reaction. Lower activation energy\ Name a substance that could 'poison' the catalysts of the catalytic converter. Lead compounds [2013 -- HL -- Section B -- Question 9] What is meant by chemical equilibrium? A state in which the rate of the forward reaction equals the rate of the reverse reaction\ Why is a chemical equilibrium described as dynamic? Reaction has not stopped.\ State Le Châtelier's principle. If a system at equilibrium is subjected to a stress (3) it tends to oppose the stress.\ (b) When a yellow solution of iron(III) chloride (FeCl~3~) and a colourless solution of potassium thiocyanate (KCNS) were mixed in a test tube, a red colour appeared and the following equilibrium was established:\ Fe^3+^ (aq) + CNS^--^ (aq) Fe(CNS)^2+^ (aq) yellow red\ Explain\ (i) the effect on the Fe^3+^ ion concentration of adding KCNS to the equilibrium mixture, Decrease -- Reaction shifts forward to oppose stress\ (ii) why changing the pressure has no effect on this equilibrium. It is an equilibrium in solution, no gases are involved\ (c) Write the equilibrium constant (Kc) expression for this reaction.\ A mixture of 1.0 × 10^--3^ moles each of iron(III) chloride and potassium thiocyanate was allowed to come to equilibrium in 1 litre of solution at room temperature according to the equation above. It was found that 1.1 × 10^--4^ moles Fe(CNS)^2+^ were present in the solution at equilibrium.\ Calculate the value of the equilibrium constant (K~c~) for the reaction.\ [\$\\frac{{\\lbrack NH}\_{{3\\rbrack}\^{2}}}{{\\lbrack N}\_{2\\rbrack}\\ \\lbrack H\_{2\\rbrack\^{3}}}\$]{.math.inline} *\ * Fe^3^ CNS^-^ Fe(CNS)^2+^ -------------------- -------------- -------------- -------------- Initial 1.0 x 10^-3^ 1.0 x 10^-3^ 0 Change -1x -1x +1x At equilibrium (L) 8.9 x 10^-4^ 8.9 x 10^-4^ 1.1 x 10^-4^ K~c~ = 1.1 × 10^--4^\ (8.9 × 10^--4^)^2^ = 138.87 \(d) The red colour faded when the test tube containing the equilibrium mixture was placed in an ice-water bath. State whether the value of K~c~ for this reaction is bigger or smaller at the lower temperature. Is the forward reaction exothermic or endothermic? Justify your answer. Smaller -- It is an Endothermic reaction so Cooling shifts the reaction in the backwards direction