A Level Chemistry 3.1.6 Chemical Equilibria PDF

A LEVEL CHEMISTRY 3.1.6 CHEMICAL EQUILIBRIA THE EQUILIBRIUM CONSTANT - Kp The Equilibrium Constant, Kp, is a quantitative measure of the position of a gaseous equilibrium using pressure (that’s what the “p” stands for). The ca...

A LEVEL CHEMISTRY 3.1.6 CHEMICAL EQUILIBRIA THE EQUILIBRIUM CONSTANT - Kp The Equilibrium Constant, Kp, is a quantitative measure of the position of a gaseous equilibrium using pressure (that’s what the “p” stands for). The calculation is the same as that for Kc (Year 12). You need: 1. A balanced equation for the reaction 2. The equilibrium concentrations of all reactants and products e.g. aA(g) + bB(g) ⇌ cC(g) + dD(g) c d p(A) = the equilibrium partial pressure of A p(C) p(D) Kp a b Units could be atm, Pa or kPa p(A) p(B) a = no. of moles from the equation Partial pressure is the pressure that a particular gas contributes to the overall pressure in a chemical system. Each gas present at equilibrium creates “pressure” by colliding with the walls of the container. The sum of the partial pressures of the gases = the total pressure Like Kc, Kp can provide a quantitative idea about where the position of the equilibrium lies: If Kp > 1.0 The equilibrium lies to the right (favours the products). The greater the value the further to right it lies. If Kp = 1.0 The equilibrium lies perfectly in the middle. If Kp < 1.0 The equilibrium lies to the left (favours the reactants). The smaller the value the further to left it lies. AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.1.6 CHEMICAL EQUILIBRIA DEDUCING PARTIAL PRESSURE AT EQUILIBRIM In exam questions, you may be asked to deduce the partial pressures of gases at equilibrium e.g. Take the following reaction: A(g) + B(g) ⇌ C(g) + D(g) 2.4 moles of A was reacted with 3.2 moles of B at 10.2atm pressure. At equilibrium, it was found that 0.6 moles of D were produced. Calculate the equilibrium constant, Kp, for the reaction. A(g) + B(g) ⇌ C(g) + D(g) Initial no. moles: 2.4 3.2 0 0 -0.6 -0.6 +0.6 +0.6 No. moles at equilibrium: 1.8 2.6 0.6 0.6 Now that we have the total no. moles of gas at equilibrium, we can find their mole fractions. Total number of moles present at equilibrium = 5.6 No. Moles 1.8 2.6 0.6 0.6 Mole Fraction = Total no. Moles 5.6 5.6 5.6 5.6 = 0.32 0.46 0.11 0.11 Now that we have the mole fractions we can calculate the partial pressures of each gas. Partial Pressure = Mole Fraction x Total Pressure (10.2 atm in this case) Partial Pressures = 3.26atm 4.69atm 1.12atm 1.12atm You now have the partial pressures to input into the Kp expression to calculate Kp. AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.1.9 RATE EQUATIONS HINTS | TIPS | HACKS Steps to calculate partial pressures in an exam question: 1. Deduce the number of moles of each gas present at equilibrium 2. Add the up to find the total number of moles present 3. Calculate the mole fraction (MF) for each gas using: No. Moles Mole Fraction = Total no. Moles TIP: The sum of the mole fractions should always = 1.0 4. Calculate the partial pressure for each gas using: Partial Pressure = Mole Fraction x Total Pressure TIP: The sum of the partial pressures should always = the total pressure 5. Insert the partial pressures into the Kp expression You may also be expected to rearrange the expression for Kp and use this to find a missing partial pressure value for a gas. You may also be asked to deduce the units of Kp. This can be done in the same way as Kc (See Year 12 Equilibria) Let’s not forget that anything related to le Chatelier’s Principle is also fair game here, so be prepared! How To Tackle Kp Questions AQA www.chemistrycoach.co.uk © scidekick ltd 2024