a.) The rate constant for the reaction is 0.460 M–1·s–1 at 200 °C. A--> products

Question

a.) The rate constant for the reaction is 0.460 M–1·s–1 at 200 °C. A--> products. If the initial concentration of A is 0.00680 M. what is the concentration after 315 s?
b.)The rate constant for this zero-order reaction is 0.0190 M·s–1 at 300 °C. A--> products. How long (in seconds) would it take for the concentration of A to decrease from 0.800 M to 0.240 M?
c.)The rate constant for this second-order reaction is 0.520 M–1·s–1 at 300 °C. A--> products .How long (in seconds) would it take for the concentration of A to decrease from 0.630 M to 0.390 M?
d.)The rate constant for this first-order reaction is 0.970 s–1 at 400 °C. A--> products. How long (in seconds) would it take for the concentration of A to decrease from 0.840 M to 0.210 M?

Explanation / Answer

a.) The rate constant for the reaction is 0.460 M–1·s–1 at 200 °C. A--> products. If the initial concentration of A is 0.00680 M. what is the concentration after 315 s?

Solution :- using the second order rate equation we can find the final concentration

1/[A]t = K* t + 1/[A]o

1/[A]t = 0.460 M-1 s-1 * 315 s + (1/0.006800]

1/[A]t = 291.9588

[A]t = 1/291.9588

[A]t = 0.003425 M

So the final concentration is 0.003425 M

b.)The rate constant for this zero-order reaction is 0.0190 M·s–1 at 300 °C. A--> products. How long (in seconds) would it take for the concentration of A to decrease from 0.800 M to 0.240 M?

Solution :-

Integrated equation for the zero order reaction is as follows

[A]t = [A]o – Kt

[0.240] = [0.800] – (0.0190 M.s-1 * t)

0.240 – 0.800 = -0.0190 *t

-0.56 = -0.0190 *t

-0.56 / -0.0190 = t

29.5 s = t

So the time needed is 29.5 s

c.)The rate constant for this second-order reaction is 0.520 M–1·s–1 at 300 °C. A--> products .How long (in seconds) would it take for the concentration of A to decrease from 0.630 M to 0.390 M?
Solution :- using the second order rate equation we can calculate the time

1/[A]t = K* t + 1/[A]o

1/0.390 = 0.520 M-1 s-1 * t + (1/0.630]

2.5641 = 0.520 * t + 1.587

2.5641 – 1.587 = 0.520 * t

0.9771 = 0.520 *t

0.9771/0.520 = t

1.88 s = t

So the time needed is 1.88 second

d.)The rate constant for this first-order reaction is 0.970 s–1 at 400 °C. A--> products. How long (in seconds) would it take for the concentration of A to decrease from 0.840 M to 0.210 M?

solution :-

using the first order rate equation

ln([A]t/[A]o) = - k*t

ln [0.210/0.840] = - 0.970 s-1 * t

-1.386 = -0.970 s-1 * t

-1.386 / -0.970 s-1 = t

1.43 s = t

So the time needed is 1.43 s

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