Question

For a first order reaction find the ratio of t90% to t10%

Answer 1

Answer:

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Explanation:

ANSWER

For t  

90%

​  

 ,a  

t

​  

=  

100

10a  

0

​  

 

​  

=  

10

a  

0

​  

 

​  

 

t  

90%

​  

=  

k

1

​  

In  

a  

0

​  

/10

a−0

​  

=  

k

In10

​  

....(1)

For t  

99%

​  

, a  

t

​  

=  

100

1

​  

a−0

t  

99%

​  

=  

K

1

​  

In  

a  

0

​  

/100

a  

0

​  

 

​  

=  

k

2In10

​  

....(2)

From (1) and (2),

t  

99%

​  

=2×t  

90%

​  

=>x=2

Answer 2

Answer:

The ratio of $t_{90\%}$ to $t_{10\%}$ is $20$.

Explanation:

Given,

The reaction occurring is the first-order reaction.

The ratio of $t_{90\%}$ to $t_{10\%}$ =?

As we know, the integrated rate law expression for the first-order reaction is:

• $k = \frac{2.303}{t} log\frac{a}{a-x}$
• $t = \frac{2.303}{k} log\frac{a}{a-x}$   -------equation (1)

Here,

• k = The rate constant
• a = Initial concentration
• x = The amount reacted
• a-x = Final concentration

Case 1) For $t_{90\%}$ , x = $90\%$

Let the initial concentration, a = $100\%$

Therefore, a-x = $100\% - 90\%$ = $10\%$

After putting the values in equation (1), we get:

• $t_{90\%} = \frac{2.303}{k} log\frac{100}{10}$
• $t_{90\%} = \frac{2.303}{k} log 10$
• $t_{90\%} = \frac{2.303}{k}$               ( $log 10 = 1$ )

 

Case 2) For $t_{10\%}$, x = $10\%$

Let the initial concentration, a = $100\%$

Therefore, a-x = $100\% - 10\%$ = $90\%$

After putting the values in equation (1), we get:

• $t_{10\%} = \frac{2.303}{k} log\frac{100}{90}$
• $t_{10\%} = \frac{2.303}{k} log \frac{10}{9}$
• $t_{10\%} = \frac{2.303}{k} (log 10 - log 9)$             ( $log\frac{x}{y} = log x - log y$ )
• $t_{10\%} = \frac{2.303}{k} (1 - 0.95)$                    ( $log 10 = 1$ , $log 9 = 0.95$ )
• $t_{10\%} = \frac{2.303}{k}\times 0.05$

Now, the ratio of $t_{90\%}$ to $t_{10\%}$ is:

• $\frac{t_{90\%}}{t_{10\%}} = \frac{\frac{2.303}{k} }{\frac{2.303\times 0.05}{k} }$
• $\frac{t_{90\%}}{t_{10\%}} = \frac{1}{0.05}$
• $\frac{t_{90\%}}{t_{10\%}} = 20$

Hence, the ratio of $t_{90\%}$ to $t_{10\%}$ is $20$.