The following reaction was monitored as a function of time: A rightarrow B + C A

Question

The following reaction was monitored as a function of time: A rightarrow B + C A plot of ln[A] versus time yields a straight line with slope -4.4 times 10-3/s. Part B Write the rate law for the reaction. Rate = k Rate = k[A] Rate = k[A]2 Rate = k[A]3 Submit My Answers Give Up Correct Part C What is the half-life? Express your answer using two significant figures. t1/2 = 160 s Submit My Answers Give Up Correct Part D If the initial concentration of A is 0.260 M , what is the concentration after 230 s ? A = Submit My Answers Give Up

Explanation / Answer

Part B

Solution:

Lets write integrated first law:

Ln [A]t = -kt + ln [A]0

y              mx        c

from this equation we say that slope = -k , and the it has straight line.

In given problem we get slope = -4.4 E-3. This indicates that the order with respect to in this reaction is first.

So ,

Rate = k [A]

Part C :

To calculate half life for first order reaction we use following equation

Half life (t ½ ) = 0.693 / k

Slope = -k

k = 4.4 E-3 per s

Half life (t ½ ) = 0.693 / 4.4 E-3 per s

Half life (t ½ )= 158 s

Answer in two sig figure = 160 s

           

Part C

Given :

Initial concentration of A = 0.260 M

t ( time ) = 230 s

Lets use integrated first law.

Ln [A]t = -kt + ln [A]0

Ln [A]t = - (4.4 E-3 per s ) x 230 s + ln (0.260)

Ln [A]t = - (4.4 E-3 per s ) x 230 s - 1.35

Ln [A]t = - (4.4 E-3 per s ) x 230 s - 1.35

Ln [A]t = -2.359

Lets take exp of bot side

[A]t = 0.0945 M

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