Question

A certain substance X, decomposes. 75% of X remains after 100 minutes. How much X remains after 200 minutes if the reaction order with respect to X is the following order: 1. Zero order, 2. First order, 3. Second order

Answer 1

From the first order reaction system we know that,

K=

t

2.303log

A

A

0

k=

100

2.303log

25

100

For 200 g,

100

2.303log

25

100

=

t

2.303log

50

200

0.602/100=0.602/t

t=100minutes

Answer 2

Answer:

1. 50% of X will remain after 200 minutes

2. 56.25% will be remaining after 200 minutes.

3. 60% will be remaining after 200 minutes.

Explanation:

Case 1: When the reaction is considered a zero-order reaction.

let the initial amount of reactant ($A_{o}$) = 100

amount of reactant remaining after 100 minutes ($A_{t}$) = 75

Time taken by the reaction (t) = 100 minutes

The rate constant of the reaction (K) =?

As we know that

$\[{A_o} - {A_t} = kt\]$

100 - 75 = 100×K

K = 0.25 $molL^{-1}s^{-1}$

Now,

let the initial amount of reactant ($A_{o}$) = 100

amount of reactant remaining after 200 minutes ($A_{t}$) = ?

Time taken by the reaction (t) = 200 minutes

The rate constant of the reaction (K) = 0.25 $molL^{-1}s^{-1}$

As we know that

$\[{A_o} - {A_t} = kt\]$

100 - $A_{t}$ = 200×0.25

100-50 = $A_{t}$

50 = $A_{t}$

Therefore, 50% of X will remain after 200 minutes

Case 2: When the reaction is considered a first-order reaction.

let the initial amount of reactant ($A_{o}$) = 100

amount of reactant remaining after 100 minutes ($A_{t}$) = 75

Time taken by the reaction (t) = 100 minutes

The rate constant of the reaction (K) =?

As we know that

$\[\begin{gathered} 2.303\log \left( {\frac{{{A_o}}}{{{A_t}}}} \right) = Kt \hfill \\ 2.303\log \left( {\frac{{100}}{{75}}} \right) = 100 \times K \hfill \\ 2.303 \times 0.12493 = 100 \times K \hfill \\ K = 2.877 \times {10^{ - 3}}{s^{ - 1}} \hfill \\ \end{gathered} \]$

Now,

let the initial amount of reactant ($A_{o}$) = 100

amount of reactant remaining after 200 minutes ($A_{t}$) =?

Time taken by the reaction (t) = 200 minutes

The rate constant of the reaction (K) = $2.877\times10^{-3}$$s^{-1}$

As we know that

$\[\begin{gathered} 2.303\log \left( {\frac{{{A_o}}}{{{A_t}}}} \right) = Kt \hfill \\ 2.303\log \left( {\frac{{100}}{{{A_t}}}} \right) = 200 \times 2.877 \times {10^{ - 3}} \hfill \\ \log \left( {\frac{{100}}{{{A_t}}}} \right) = 0.249877 \hfill \\ \frac{{100}}{{{A_t}}} = {10^{0.249877}} \hfill \\ {A_t} = 56.25 \hfill \\ \end{gathered} \]$

Therefore, 56.25% will be remaining after 200 minutes.

Case 3: When the reaction is considered a Second-order reaction.

let the initial amount of reactant ($A_{o}$) = 100

amount of reactant remaining after 100 minutes ($A_{t}$) = 75

Time taken by the reaction (t) = 100 minutes

The rate constant of the reaction (K) =?

As we know that

$\[\begin{gathered} \frac{1}{{{A_t}}} - \frac{1}{{{A_o}}} = Kt \hfill \\ \frac{1}{{75}} - \frac{1}{{100}} = K \times 100 \hfill \\ \frac{{100 - 75}}{{75 \times 100}} = K \times 100 \hfill \\ \frac{{25}}{{7500 \times 100}} = K \hfill \\ K = 3.33 \times {10^{ - 5}}Lmo{l^{ - 1}}{s^{ - 1}} \hfill \\ \end{gathered} \]$

Now,

let the initial amount of reactant ($A_{o}$) = 100

amount of reactant remaining after 200 minutes ($A_{t}$) =?

Time taken by the reaction (t) = 200 minutes

The rate constant of the reaction (K) = $\[3.33 \times {10^{ - 5}}Lmo{l^{ - 1}}{s^{ - 1}}\]$

As we know that

$\[\begin{gathered} \frac{1}{{{A_t}}} - \frac{1}{{{A_o}}} = Kt \hfill \\ \frac{1}{{{A_t}}} - \frac{1}{{100}} = 3.33 \times {10^{ - 5}} \times 200 \hfill \\ \frac{1}{{{A_t}}} = 3.33 \times {10^{ - 5}} \times 200 + \frac{1}{{100}} \hfill \\ {A_t} = 60 \hfill \\ \end{gathered} \]$

Therefore, 60% will be remaining after 200 minutes.