Download Chemical Equilibrium 2 - Principles of Chemistry - Lecture Slides | CEM 152 and more Study notes Chemistry in PDF only on Docsity! Chpt 15 Chemical Equilibrium Calculating Equilibrium Constants Using the equilibrium concentrations of all reactants and products, K can be calculated. Typically only a few concentrations are known and stoichiometry must be used to fill in the blanks. At 500 oC the Haber process was started with 0.93136 atm of H2 and 0.43312 atm of N2. At equilibrium there were 0.928 atm of H2. What is Kp? CEM 152 – SS2011 N2(g) 3 H2(g) 2 NH3(g) Initial 0.43312 0.93136 Change Equilibrium ? 0.928 ? Chpt 15: Equilibria Le Chatelier’s Principle Le Chatelier’s Principle states that if an equilibrium system is disturbed the system will shift to counteract the effect of the disturbance. Disturbing the system involves CEM 152 – SS2011 Adding or removing reactants or products Changing the pressure Changing the temperature Chpt 15: Equilibria Le Chatelier’s Principle: Concentration When a substance is added the equilibrium will shift to consume some of the added substance. If NH3 is removed equilibrium is disturbed and more NH3 is produced. This enables more NH3 to be produced from the same reactants. If either N2 or H2 were added more NH3 would also be produced. If H2 were removed some NH3 would be destroyed CEM 152 – SS2011 322 2NHH3N Chpt 15: Equilibria Le Chatelier’s Principle: Pressure If the volume is decreased the partial pressures of each of the gases are increased. Pressure increases forces the reaction to NH3 since there are fewer product gas molecules. However, increasing the total pressure of the system by adding a gas that does not react will not affect the equilibrium. CEM 152 – SS2011 322 2NHH3N Chpt 15: Equilibria Catalysis A catalyst lowers the activation energy for both the forward and reverse reaction. A catalyst will increase the rate at which a reaction obtains equilibrium. not change the position of the equilibrium. In the Haber process, the temperature is increased to provide an acceptable reaction rate (at the expense of lower Kp). If a catalyst were developed allowing the reaction to proceed at lower temperature the economic savings would be significant. CEM 152 – SS2011Chpt 15: Equilibria For the exothermic Haber-Bosch process N2 + 3H2 → 2NH3 which of the following conditions would not result in the production of more NH3. A – add N2 B – add H2 C – remove NH3 D – decrease T E – none of the above Chpt 15: Equilibria CEM 152 – SS2011 Reaction Coordinate Question The reaction coordinate diagram represents the conversion from reactants to products with two different activation energies, solid line and dotted line. What is the relationship between the equilibrium constants? A – Ksolid > Kdotted B – Ksolid < Kdotted C – Ksolid = Kdotted D – not enough information Question Chpt 15: Equilibria CEM 152 – SS2011 reactants products
