Consider the 1^st order decomposition of hydrogen peroxide and use the graph bel

Question

Consider the 1^st order decomposition of hydrogen peroxide and use the graph below to answer the question that follow. The data in the graph was collected at 298k. 2H_2 O_2 (aq) rightarrow 2H_2 O + O_2 (g) What is the numerical value of the rate constant for this 1^st order reaction? What would the rate be if the concentration of H_2 O_2 were 1.30M How long would it take the concentration to decrease from 0.6M to 0.07M at 298K? If you ra3ise the temperature to 325K, you find the rate of the reaction has increased by a factor of 1.9 What is the activation energy for this reaction?

Explanation / Answer

1)

[Ro] = 1 M

at t = 1100 s , [R] =0.4 M

use:

[R] = [Ro]*e^(-k*t)

0.4 = 1 *e^(-k*1100)

k = 8.33*10^-4 s-1

Answer: 8.33*10^-4 s-1

2)

rate law can be written as:

rate = k*[H2O2]o

old rate = 8.33*10^-4*1 M

= 8.33*10^-4 M/s

rate new / rate old = [H2O2]new / [H2O2]old

rate new / 8.33*10^-4 = 1.30/1

new rate = 1.08*10^-3 M/s

Answer: 1.08*10^-3 M/s

3)

[R1] = [Ro]*e^(-k*t1)

[R2] = [Ro]*e^(-k*t2)

so,

[R2] / [R1] = e^(-k*(t2-t1))

0.07/0.6 = e^(-8.33*10^-4 (t2-t1))

t2-t1 = 2579 s

= 43minutes

Answer: 43 minutes

4)

use:

ln (K2/K1) = Ea/R * (1/T1 - 1/T2)

ln(1.9) = Ea/8.314 * (1/298 - 1/325)

0.642= Ea/8.314 * 2.788*10^-4

Ea =19146 J/mol

= 19.15 KJ/mol

Answer: 19.15 KJ/mol

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