Question

In a zero order reaction, the time taken to reduce the concentration of reactant from
50% to 25% is 30 minutes. What is the time required to reduce the concentration from
12.5% to 6.25%​

Answer 1

Answer:

answer is 7.81 u do not have option so i cant sure 100 percent

Explanation:

in zero order

A0-At=kt

so;A0-At/t=k

50/100=0.5

25/100=0.25

0.5-0.25/30=0.008

then again 12.5/100=0.125

6.25/100=0.0625

so using A0-At/k=t then

0.125-0.0625/0.008=7.81

Answer 2

The time required to reduce the concentration from  12.5% to 6.25%​ will be 7.5 minutes

Explanation:

Integrated rate law for zero order kinetics is given as:

$k=\frac{1}{t}([A_o]-[A])$

k = rate constant of the reaction = ?

$[A_o]$ = Concentration of reactant initially 50%  = 50

$A$ = Concentration of reactant at time = 25 % = 25

t = time = 30 min

$k=\frac{1}{30}([50]-[25])$

$k=0.83min^{-1}$

$t=\frac{1}{0.83}([12.5]-[6.25])$

$t=7.5min$

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