Chemical Equilibria 2 Lecture Notes PDF

MED-102 General Chemistry Chemical Equilibrium 2  LOBs covered Perform various types of calculations on chemical systems at equilibrium Predict what will happen to a chemical reaction if it is disturbed at equilibrium in various ways Discuss the interplay between chemical...

MED-102 General Chemistry Chemical Equilibrium 2  LOBs covered Perform various types of calculations on chemical systems at equilibrium Predict what will happen to a chemical reaction if it is disturbed at equilibrium in various ways Discuss the interplay between chemical kinetics and chemical equilibrium  Using the Equilibrium Expression Calculating equilibrium concentrations 2 NO(g) + O2(g) 2 NO2(g) Kc = 6.9 × 105 at 500 K [O2 ] = 1.0 10−3 M, [NO2 ] = 5.0 10−2 M [NO 2 ]2 [NO 2 ]2 Kc = 2 , so that [NO]= [NO] [O 2 ] [O 2 ]K c (5.0 10−2 ) 2 −6 −3 [NO]= −3 = 3.6  10 = 1.9  10 M (1.0 10 )(6.9 10 )5  Using the Equilibrium Expression Calculating equilibrium concentrations H2(g) + I2(g) 2 HI(g) Kc = 57.0 at 700 K t = 0 0.100 M 0.100 M 0 t =  0.100 – x 0.100 – x 2x [HI]𝟐 (𝟐𝒙)𝟐 𝑲𝒄 = = = 𝟓𝟕. 𝟎 [H𝟐 ][I𝟐 ] (𝟎. 𝟏𝟎𝟎 − 𝒙)(𝟎. 𝟏𝟎𝟎 − 𝒙) x = 0.079 M [H2] = [I2] = 0.100-0.079 = 0.021 M and [HI] = 2(0.079) = 0.158 M  Quadratic Equation or not? [HI]𝟐 (𝟐𝒙)𝟐 𝑲𝒄 = = = 𝟓𝟕. 𝟎 [H𝟐 ][I𝟐 ] (𝟎. 𝟏𝟎𝟎 − 𝒙)(𝟎. 𝟏𝟎𝟎 − 𝒙) Should we expand this expression into a quadratic equation? −b  b 2 − 4ac ax 2 + bx + c = 0 x1,2 = 2a Not if we don’t have to. (2 x) 2 (2 x) 2 2x K c = 57.0 = K c = 57.0 = 7.55 = = (0.100 − x) 2 (0.100 − x) 2 0.100 − x  Using the Equilibrium Expression – Revision Slide [HI]𝟐 (𝟐𝒙)𝟐 𝑲𝒄 = = = 𝟓𝟕. 𝟎 [H𝟐 ][I𝟐 ] (𝟎. 𝟏𝟎𝟎 − 𝒙)(𝟎. 𝟏𝟎𝟎 − 𝒙) Solving this equation for x seems difficult. If we expand the entire equation out, we will end up with a quadratic equation of the type ax2 + bx + c = 0, which has two possible solutions that must be obtained through the quadratic equation. This is somewhat complicated. However, for this particular system, we notice that the two terms in the denominator are both the same, (0.100 – x). So, the denominator is actually (0.100 – x)2. This enables us to take the square roots of both sides, ending up with 2x/(0.100 – x) = (57.0)1/2. This system is now linear in x, and there are no terms in x2. It is relatively simple to solve. This way, we can obtain x. Remember, however, x is the change in the concentration, and we must go back to the initial equilibrium setup to find the equilibrium concentrations. WATCH: https://www.youtube.com/watch?v=3O2QGD1vPJA  Le Châtelier’s Principle Le Châtelier’s Principle: if a stress is applied to a system that is at equilibrium, net reaction occurs in the direction that relieves the stress It is important to be able to manipulate the position of an equilibrium so as to make as much product as possible using as less time (energy) as possible Haber Process  Haber Process – Revision Slide The Haber process is the industrial process for making ammonia from its elements. We mix together 1 part of N2 and 3 parts of H2 (see balanced equation). We allow them to react, reaching equilibrium. The reaction conditions indicate a very high pressure (200 atm) and a medium temperature (400-450 oC). An iron catalyst is also used. When equilibrium is reached, we cool the equilibrium gas mixture down. Since NH3 is polar but H2 and N2 are nonpolar, the condensation point (boiling point) of NH3 is the highest and will be reached first. This will turn NH3 into a liquid, and it can be easily removed from the reaction container. The unreacted H2 and N2 gases are recycled, giving more NH3. This is repeated until they all react.  Changes in Concentration Haber process [NH3 ]2 Qc = = Kc 3 [H 2 ] [N 2 ] Adding N2 reduces the value of Qc. To increase it again up to Kc, we make more NH3, using up the H2 and N2. Removing the NH3 reduces the value of Qc. This forces the system to make more NH3. Therefore, it makes sense to continuously remove NH3 to force the system to use up all the reactants and to maximize the amount of NH3 formed.  Changes in Concentration Haber process – adding hydrogen gas The final concentrations are different than the initial, BUT the value of the equilibrium constant does not change provided the temperature stays constant. 5-Minute Break  Changes in Pressure and Volume Haber process Reducing the volume increases the pressure. pV = nRT so P = nRT/V According to Le Chatelier’s Principle, the system has to do something to reduce the pressure. Reduce P by decreasing n (moles of gas) P = nRT/V Left side = 4 moles Right side = 2 moles Therefore, the system goes to the right side, which has fewer moles of gas. This forces the system to make more NH3. This is why we use a high pressure for the Haber process. Changes in Pressure and Volume What about these systems? H2(g) + I2(g) 2 HI(g) C(s) + H2O(g) CO(g) + H2(g)  Changes in Pressure and Volume – Revision Slide H2(g) + I2(g) 2 HI(g) For this system, the number of moles of gas is the same on both sides. This means that changing the pressure will have absolutely no effect on the position of the equilibrium. C(s) + H2O(g) CO(g) + H2(g) For this system, we have one mole of gas on the left side, and 2 moles of gases on the right side. If we increase the pressure on this system, the equilibrium will shift to the left, where we have fewer moles of gas. Therefore, if we want to make more products, we need to lower the pressure.  Changes in Temperature Haber Process Ho = -92.2 kJ + 92.2 kJ For an exothermic process, we consider heat as another product. If we raise the temperature, the system will perceive it as input of heat (product). It will shift to the left side. N2O4(g) 2 NO2(g) Ho = +55.3 kJ N2O4(g) + 55.3 kJ 2 NO2(g) For an endothermic process, we consider heat as a reactant. It must be added to the reactants side. Therefore, increasing the temperature here is like adding more reactant. The equilibrium will shift to the right side.  Effect of a Catalyst A catalyst speeds up reaction rate Does it affect equilibrium position? NO, because the forward and reverse reaction rates are affected to the same degree  Kinetics versus Equilibrium The Haber process is exothermic. Therefore, increasing the temperature will shift the equilibrium position to the left, making less NH3. However, decreasing the temperature to push the equilibrium to the right, will slow down the reaction. So, you will make more ammonia, but a lot slower. This is not good for an industrial process. A slower process will need more time and energy to complete. Therefore, we use a medium temperature as a compromise between the equilibrium position and the rate of reaction.  Le Châtelier’s Principle – Revision Slide Now that we have covered the topic, let me invite you to go to YouTube and watch a really clear video about Le Châtelier’s Principle: WATCH: https://www.youtube.com/watch?v=XmgRRmxS3is  Summary for Revision Another way to use equilibrium constants is in determining equilibrium concentrations. Two examples of such calculations have been provided in the presentation. Le Châtelier’s Principle states that if a system at equilibrium is placed under a stress, it will shift the equilibrium position in such a way as to reduce the stress. If more reactant is added to a system at equilibrium, the equilibrium will shift to the right to use part of it up, forming more product. If reactant is removed from the equilibrium mixture, the system will shift to the left to make more of it and re-establish equilibrium. If more product is added to the equilibrium mixture, the system will shift to the left to use part of it up. If product is removed from the equilibrium mixture, the system will more to the right to make more of it and re-establish equilibrium. If the pressure of an equilibrium mixture is increased, the system will move to the side with the fewest number of moles of gas no reduce the pressure. If the pressure is decreased, the system will shift to the side with more moles of gas to re-establish equilibrium. In an exothermic system, increasing the temperature shifts the system to the left. Reducing the temperature shifts the system to the right to make more product. In an endothermic system, increasing the temperature shifts the system to the right. Reducing the temperature shifts the system to the left. A catalyst does not affect the position of the equilibrium, only the rate of the reaction.