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3 years ago

10

Cyclopropane is converted to propene in a first–order process with a rate constant of 5.4 × 10–2 hour–1 . The initial concentrat

ion of cyclopropane is 0.150 M. Estimate the concentration of cyclopropane after 22.0 hours

1 answer:

Hitman42 [59]3 years ago

3 0

Answer:

The concentration of cyclopropane after 22.0 hour is 0.0457 M.

Explanation:

Conversion of cyclopropane into propene follows first order kinetics.

The integrated rate of first order kinetic is given by :

$[A]=[A_o]\times e^{-kt}$

$[A_o]$ = Initial concentration of reactant

$[A]$ = final concentration of reactant after time t

k = rate constant of the reaction

We have :

Rate constant of the reaction = k = $5.4\times 10^{-2} hour^{-1}$

$[A_o]=0.150 M$

t = 22.0 hour

[A] =?

$[A]=0.150 M\times e^{-5.4\times 10^{-2} hour^{-1}\times 22.hour}$

$[A]=0.0457 M$

The concentration of cyclopropane after 22.0 hour is 0.0457 M.

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3 years ago

Since the half-life of 235U (7. 13 x 108 years) is less than that of 238U (4.51 x 109 years), the isotopic abundance of 235U has

Ymorist [56]

Answer:

$\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

Explanation:

Given that:

The Half-life of $^{235}U$ = $7.13 \times 10^8 \ years$ is less than that of $^{238} U = 4.51 \times 10^9 \ years$

Although we are not given any value about the present weight of $^{235}U$ .

So, consider the present weight in the percentage of $^{235}U$ to be y%

Then, the time elapsed to get the present weight of $^{235}U$ = $t_1$

Therefore;

$N_1 = N_o e^{-\lambda \ t_1}$

here;

$N_1$ = Number of radioactive atoms relating to the weight of y of $^{235}U$

Thus:

$In( \dfrac{N_1}{N_o}) = - \lambda t_1$

$In( \dfrac{N_o}{N_1}) = \lambda t_1$ --- (1)

However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of $^{235}U$ to be = $t_2$

Then:

$In( \dfrac{N_o}{N_2}) = \lambda t_2$ ---- (2)

here;

$N_2$ = Number of radioactive atoms of $^{235}U$ relating to 3.0 a/o weight

Now, equating equation (1) and (2) together, we have:

$In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) = \lambda( t_1-t_2)$

replacing the half-life of $^{235}U$ = $7.13 \times 10^8 \ years$

$In( \dfrac{N_2}{N_1}) = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)$ ( since $\lambda = \dfrac{In 2}{t_{1/2}}$ )

∴

$\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}$

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o

Thus, The time elapsed is $\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

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