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Chrome Secure https://session.ma tenngchemistry.com/myct/itemView?assignmentProblemlD-92076515 with Feedback Exercise 14.54 -E Exercise 14.54 Enhanced with Feedback reaction was monitored as a function AB A + B A plot of 1/[AB] versus time yields a straight line with slope 5.2-102 (M-s) Part c What is the half-life when the initial concentration is 0.58 M? Express your answer using two significant figures. Subrnit Incorrect; Try Again; 5 attempts remaining Part D If the initial concentration of AB is 0 290 M, and the reaction mixture initia concentrations of A and B after 80 s? Express your answers numerically using two significant figures, sepa

Explanation / Answer

Ans. Given, a plot of 1/[AB] versus time yields a straight line with slope 5.2 x 10-2 M-1 s-1.

If 1/[concertation of reactants] vs time yields a straight line, it is a second order reaction.

The slope of 1/[concertation] vs time graph is equal to the rate constant.

Therefore, rate constant of the reaction, k = slope = 5.2 x 10-2 M-1 s-1.

Part C: The half-life of a second order reaction is given as-

                        t½ = 1 / k[A]0           - equation 1

Where, [A]0 = Initial concentration of reactant (here, AB).

Note: Don’t confuse [A]0 with the product A of the reaction. Here, [A]0 = [AB].

k = rate constant

Putting the values in above equation –

                        t½ = 1 / [5.2 x 10-2 M-1 s-1 x 0.58 M] = 1 / 0.03016 s-1 = 33.156 s

Thus, half-life of the reaction at given [AB] = 33.156 s

Part D: Using second order kinetics-

1/ [A] = kt + (1/ [A]0)        - equation 1

Where, [A]0 = Initial concentration of reactant (here, AB).

Note: Don’t confuse [A]0 with the product A of the reaction. Here, [A]0 = [AB].

[A] = Final concentration of reactant after time t

k = rate constant

t = time of reaction.

Putting the values in above equation-

            1/ [AB]t = (5.2 x 10-2 M-1 s-1) x 80s + (1 / 0.290 M)

            Or, 1/ [AB]t = 4.160 M-1 + 3.448 M-1 = 7.608 M-1

            Or, [AB]t = 1 / (7.608 M-1) = 0.131 M

Therefore, remaining [AB] after 80s = 0.130.

# Amount of AB consumed during reaction = Initial [AB] – Remaining [AB] after 80s

                                                            = 0.290 M – 0.131 M

                                                            = 0.159 M

# According to the stoichiometry of balanced reaction, 1 mol AB produces 1 mol each of A and B.

Now,

Amount of [A] formed = Amount of [B] formed = Amount of [AB] consumed = 0.159 M

Therefore, after 80s-

            [A] = 0.16 M

            [B] = 0.16 M

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