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Savatey [412]
3 years ago
7

A second order reaction is 50% complete in 15 min. How long after the start of the 10 reaction will it be 85% complete?

Chemistry
1 answer:
DiKsa [7]3 years ago
3 0

Answer: 85 minutes

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

Half life for second order kinetics is given by:

t_{\frac{1}{2}=\frac{1}{k\times a_0}

t_{\frac{1}{2} = half life = 15 min

k = rate constant =?

a_0 = initial concentration = 100 (say)

15min=\frac{1}{k\times 100}

k=\frac{1}{1500}

Integrated rate law for second order kinetics is given by:

\frac{1}{a}=kt+\frac{1}{a_0}

a= concentration left after time t = 100-\farc{85}{100}\times 100=15

\frac{1}{15}=\frac{1}{1500}\times t+\frac{1}{100}

t=85min

Thus after 85 minutes after the start of the reaction, it will be 85% complete.

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<u>Explanation:</u>

We are given:

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<u>Initial:</u>               2.4       3.9

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So, equilibrium partial pressure of methane gas = (2.4 - x) = [2.4 - 2.167] = 0.233 atm

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The expression of K_p for above equation follows:

K_p=\frac{p_{CO}\times (p_{H_2})^3}{p_{CH_4}\times p_{H_2O}}

Putting values in above equation, we get:

K_p=\frac{2.167\times (6.5)^3}{0.233\times 1.733}\\\\K_p=1473.8

Hence, the pressure equilibrium constant for the reaction is 1473.8

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