Ch04 Chemical Changes

CHEMISTRY 1ST PAPER CHAPTER 4 Chemical Changes Question 1 A2B4(g) Δ −−−⇀ ↽−−−2 AB2(g) Here, the atomic numbers of A and B are 7 and 8. At 25 ◦C and 1 atm pressure, the degree of dissociation of A2B4 is 15%. (a) What are quantum numbers? (b) What are the bonds present in [Ag(NH3)2]Cl compound? Explain. (c) As per stem reaction, establish the relation between 𝐾𝑝and 𝐾𝑐of it. (d) As per stem, whether the degree of dissociation of A2B4 will be increased by 5% if the present pressure is reduced to 0.65 atm? Explain mathematically. Solution (a) Quantum numbers are the four numbers that describe an electron in an atom. They give its energy level, orbital shape, orientation and spin. (b) The compound has the complex ion [Ag(NH3)2]+ and the ion Cl–. Three types of bonds are present. Bond Position Number Ionic bond between [Ag(NH3)2]+ and Cl– 1 Coordinate covalent bond N −−−→Ag, from each NH3 2 Covalent bond N–H in each NH3 6 Each nitrogen atom donates its lone pair to an empty orbital of Ag+. (c) A is nitrogen and B is oxygen. The reaction is N2O4(g) −−−⇀ ↽−−−2 NO2(g) In terms of molar concentrations and partial pressures, 𝐾𝑐= [NO2]2 [N2O4] 𝐾𝑝= 𝑃2 NO2 𝑃N2O4 For 𝑛moles of an ideal gas in volu

Here 𝐶is the molar concentration of the gas. Hence 𝑃NO2 = [NO2] 𝑅𝑇and 𝑃N2O4 = [N2O4] 𝑅𝑇. 𝐾𝑝= ([NO2] 𝑅𝑇)2 [N2O4] 𝑅𝑇 = [NO2]2 [N2O4] × (𝑅𝑇)2−1 = 𝐾𝑐(𝑅𝑇)1 The change in the number of gas moles is Δ𝑛= 2 −1 = 1. 𝐾𝑝= 𝐾𝑐(𝑅𝑇)Δ𝑛= 𝐾𝑐𝑅𝑇 (d) Let the initial amount of N2O4 be 1 mol. Its degree of dissociation is 𝛼and the total pressure is 𝑃. N2O4(g) 2 NO2(g) Initial moles 1 0 Moles at equilibrium 1 −𝛼 2𝛼 Partial pressure 1 −𝛼 1 + 𝛼𝑃 2𝛼 1 + 𝛼𝑃 Total moles at equilibrium = (1 −𝛼) + 2𝛼= 1 + 𝛼. 𝐾𝑝= 𝑃2 NO2 𝑃N2O4 =  2𝛼 1 + 𝛼𝑃 2 1 −𝛼 1 + 𝛼𝑃 = 4𝛼2𝑃 1 −𝛼2 At 1 atm. Here 𝛼= 0.15 and 𝑃= 1 atm. 𝐾𝑝= 4 × (0.15)2 × 1 1 −(0.15)2 atm = 0.09 0.9775 atm = 0.0921 atm At 0.65 atm. The temperature is the same. Hence 𝐾𝑝stays constant. Let the new degree of dissociation be 𝛼′. 4𝛼′2 × 0.65 1 −𝛼′2 = 0.0921 2.6 𝛼′2 = 0.0921 −0.0921 𝛼′2 𝛼′2 = 0.0921 2.6921 = 0.03421 𝛼′ = 0.185 The new degree of dissociation is 18.5%. Increase = 18.5% −15% = 3.5% A fall of pressure favours the side with more gas moles. Hence the dissociation increases. The increase is only 3.5%. Therefore the degree of dissociation will not be increased by 5%. ahbab.dev 2

Question 2 A2(g) and B2(g) gas mixture at 500 ◦C and 50 atm pressure reacts, and at equilibrium the mixture contains 25% AB3(g) together with producing 92 kJ heat. Reaction progress Potential energy A2 + 3 B2 2 AB3 Δ𝐻= −92 kJ (a) Write law of mass action. (b) What is blanching in food preservation? (c) Calculate the value of 𝐾𝑝of the given stem reaction. (d) How can you produce maximum amount of product applying Le Chatelier principle? Discuss it. Solution (a) The rate of a reaction is proportional to the active masses of the reactants. The temperature must stay constant. (b) Blanching is a short heating of vegetables or fruits before they are preserved. The food is kept in boiling water or steam for one to three minutes. It is then cooled quickly in cold water. The heat destroys the enzymes and many microorganisms on the surface. Hence the colour, taste and food value stay good during storage. (c) The reaction is A2(g) + 3 B2(g) −−−⇀ ↽−−−2 AB3(g). The total pressure is 𝑃= 50 atm. The mixture contains 25% AB3. The remaining 75% is A2 and B2. They react in the ratio 1 : 3 and are taken in the same ratio. A2 B2 AB3 Mole fraction 0.75 × 1 4 = 0.1875 0.75 × 3 4 = 0.5625 0.25 Partial pr