can I have help with 1, 2 and 5 please. 67 16 Determination of a Keq Practice Qu

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can I have help with 1, 2 and 5 please.

67 16 Determination of a Keq Practice Questions 1 A 0.9% (w/w)aque solution? What is the molality of the solution in #1? Which of the following concentration units is not sensitive to temperature: and/or percent by mass? From the tabl (include the proper units on k): ous NaCI solution has a density of 1.0 g/ml. What is the molarity of this 2, 3. molarity, molality, 4. e of information below determine the rate law for A and B in the generic reaction Initial [AI (M) Initial [B] (M) Rate (M/s) 0.25 0.50 0.25 0.40 0.40 1.20 1.32x10-2 2.55x10-2 0.120 as written? Explain. Introduction and Objectives The equilibrium constant is a thermodynamic quantity that allows the determination of the extent of a chemical reaction. Using the equilibrium constant it is possible to calculate the concentrations of reactants and products present at equilibrium for a particular chemical reaction at a particular temper reaction of interest here is: ature. The The product (a complex ion) formed in this reaction is intensely colored red. Its concentration will be rotions uill he determined using simple

Explanation / Answer

Ans. #1. Given, [NaCl] = 0.9% (w/w). That is, there is 0.9 g NaCl per 100.0 g of solution.

                        Density of solution = 1.0 g/ mL

Now, on a per 100.0 g solution basis-

Moles of solute (NaCl) = Mass / Molar mass = 0.9 g/ (58.44 g/ mol) = 0.0154 mol

Mass of solvent = Mass of solution – Mass of solute

                                    = 100.0 g – 0.9 g

                                    = 99.1 g

                                    = 0.0991 kg

# Molality of solution = Moles of solute / mass of solvent in kg

                                    = 0.0154 mol / 0.0991 kg

                                    = 0.1554 mol/ kg

                                    = 0.1554 m

#2. So far, we have-

Moles of NaCl = 0.0154 mol

Volume of solution = Mass / density

= 100.0 g / (1.0 g/ mL) = 100.0 mL = 0.100 L

# Now,

Molarity = Moles of solute / Volume of solution in liters

                                    = 0.0154 mol / 0.100 L

                                    = 0.154 mol/ L

                                    = 0.154 M

#5. Order of reaction, n with respect to [A] – comparing relative [A] and reaction rates in experiment 2 and 2; note that [B] is kept constant in both experiments-

            ([A]Rxn2 / [A]Rxn1)n = r2 / r1

            Or, (0.50 / 0.20)n = (2.55 x 10-2 / 1.32 x 10-2)

            Or, (2)n = 1.93 = 2.0 (approx..)

            Or, n = 1

Therefore, order of reaction with respect to [A] = 1

# Order of reaction, n with respect to [B] – comparing relative [B] and reaction rates in experiment 1 and 3; note that [A] is kept constant in both experiments-

            ([B]Rxn3 / [B]Rxn1)n = r3 / r1

            Or, (1.20 / 0.4)n = (0.120 / 1.32 x 10-2)

            Or, (3)n = 9

            Or, n = 2

Therefore, order of reaction with respect to [B] = 3

# Overall order of reaction = n (for [A]) + n (for [B]) = 1 + 3 = 4

Conclusion: NO, the reaction won’t occur in one single step because the theoretical stoichiometry of reaction (as shown in reaction – 1 mol A reacts with 1 mol B) is NOT equal to the molecularity of reaction (1 mol A reacts with 3 mol B). The reaction is likely to be complex and a multistep reaction.

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