Chemistry: Typical audience for a vaudeville show

Typical audience for a vaudeville show - id6114668, kaylailkanic1487

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rodriguezangelinaado

12.01.2022

half life = 1.1*10^3min , 2.24*10^3min are taken for the concentration to decrease to 25% and 4.48*10^3min for the concentration to decrease to 6.25% .

Explanation: The given information says, the reaction is first order with respect to N_2O_5 . For first order reaction, rate constant and half life are related to each other as:

half life = $\frac{0.693}{k}$

(where k stands for rate constant)

Value of k is given as 6.2*10^-^4min^-^1

half life = $\frac{0.693}{6.2*10^-^4min^-^1}$

half life = 1117.74 min or 1.1*10^3min

Let's say the initial concentration is 100. It asks to calculate the time taken to decrease the concentration to 25% which will be 25 as we have taken the initial concentration 100.

The equation that we use is:

lnA=lnA_0-kt

Where, A_0 is initial and A is remaining concentration or amounts. k is rate constant and t is the time. Let's plug in the values and do calculations for t.

ln25=ln100-6.2*10^-^4min^-^1*t

3.22=4.61-6.2*10^-^4min^-^1*t

3.22-4.61=-6.2*10^-^4min^-^1*t

-1.39=-6.2*10^-^4min^-^1*t

$t=\frac{1.39}{6.2*10*^-^4min^-^1}$

t = 2241.94 min or 2.24*10^3min

We can calculate the time when the concentration decreases to 6.25% of its initial value same as we did for the above.

ln6.25=ln100-6.2*10^-^4min^-^1*t

1.83=4.61-6.2*10^-^4min^-^1*t

1.83-4.61=-6.2*10^-^4min^-^1*t

-2.78=-6.2*10^-^4min^-^1*t

$t=\frac{2.78}{6.2*10*^-^4min^-^1}$

t = 4483.87 min or 4.48*10^3min

So, the answers for all the three parts are: half life = 1.1*10^3min , 2.24*10^3min are taken for the concentration to decrease to 25% and 4.48*10^3min for the concentration to decrease to 6.25% .

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Typical audience for a vaudeville show

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