Question

99% of a first order reaction completed in 330 minutes. Then it's t 1/2 is

Answer 1

Answer:

Step-by-step explanation:

k=  

t

2.303

​  

log  

100−x

100

​  

 

Substituting the values,

k = \dfrac{2.303}{32} log \dfrac{100}{1}k=  

32

2.303

​  

log  

1

100

​  

 

k = 0.144\space {min}^{-1}k=0.144 min  

−1

 

t = \dfrac{2.303}{0.144} log \dfrac{100}{0.1}t=  

0.144

2.303

​  

log  

0.1

100

​  

 

t = 48\space mint=48 min

Answer 2

Answer:

The correct answer is 50 minutes.

Step-by-step explanation:

90% signifies that 90% of the initial concentration gets transformed into the final product

Suppose,

Initial concentration

99% of a

$\frac{ax9}{100}$= 0.99 a = x (reacted amount)

left amount = Initial - Reacted

                  = a - 0.99a = 0.01 a

For first order

                     H$\frac{2.303}{t}$ log $\frac{a}{a-x} = \frac{2.303}{t} = log \frac{a}{0.01 a}$

Rate constant = H = $\frac{2.303}{330} log \frac{100}{1}$

                      = $\frac{2.303}{330}$ x 2 = 0.0139 = K

$\frac{t}{2} = \frac{ 0.693}{0.0139} = 49.85 =$ 50 minute (Approx)

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