16. Consider the following hypothetical aqueous reaction: A( aq )B( aq ). A flas

Question

16. Consider the following hypothetical aqueous reaction: A(aq)B(aq). A flask is charged with 0.065 mol of A in a total volume of 1000 mL. The following data are collected:

Time (min)

0

10

20

30

40

    [A] (M)

0.065

0.051

0.042

0.036

0.031

What is the order of the reaction?
(a) 0th     (b) 1st      (c) 2nd      (d) 3rd

What is the rate constant for the reaction from the data?
(a) 0.42 M1min1      (b) 0.28 M1min1      (c) 0.22 M1min1      (d) 0.18 M1min1

19. A solution of radioactive phosphorus converts to products with a first order half-life of 14.3 days. If a sample initially contains 1.530 M of the radioactive phosphorus, how much will remain after 23.0 days?

        (a) 0.328 M         (b) 0.415 M           (c) 0.502 M        (d) 0.565 M           (e) 0.822 M      (f) 0.951 M

Time (min)

0

10

20

30

40

    [A] (M)

0.065

0.051

0.042

0.036

0.031

Explanation / Answer

(16)

Let us assume the reaction is first order,

first order integrated rate constant equation is,

k = (1/t)*ln[A]0/[A]t

Case(i) After10 min, k = (1/10)*ln(0.065/0.051) = 0.0242 min-1

Case(ii) After 20 min, k = (1/20)*ln(0.065/0.042) = 0.0218 min-1

Case(iii) After 30 min, k = (1/30)*ln(0.065/0.036) = 0.0197 min-1

Since we are not getting equal rate constant values, the reaction is not first order.

Now, let us the second order reaction,

Second order integrated rate constant equation is,

1/[A]t = kt + 1/[A]0

Case (i) After 10 min, 1/0.051 = k(10) + 1/0.065 => k = 0.42 L/mol.min

Case (ii) After 20 min, 1/0.042 = k(20) + 1/0.065 => k = 0.42 L/mol.min

Case (iii) After 30min, 1/0.036 = k (30) + 1/0.065 => k = 0.42 L/mol.min

Since we are getting equal k values,

The givne reaction is second order reaction. ANd k Value is 0.42 M-1/min

(a) C and (b) A

(19)

Relation between rate constant and half life time is,

k = 0.693 / t1/2

k = 0.693 / 14.3

k = 0.0485 day-1

Now, first order integrated rate constant equation is,

k = (1/t)*ln[A]0/[A]t

0.0485 = (1/23.0)*ln(1.530/[A]t)

ln(1.530/[A]t) = 1.12

1.530 / [A]t = 3.06

[A]t = 0.500 M

(C)

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