Respuesta :
The integrated rate law expression for a first order reaction is
[tex]ln\frac{[A_{0}]}{[A_{t}]}=kt[/tex]
where
[A0]=100
[At]=6.25
[6.25% of 100 = 6.25]
k = 9.60X10⁻³s⁻¹
Putting values
[tex]ln\frac{100}{6.25}=9.6X10^{-3}t[/tex]
taking log of 100/6.25
100/6.25 = 16
ln(16) = 2.7726
Time = 2.7726 / 0.0096 = 288.81 seconds
Answer:
In 4.81 minutes the concentration of reactant will drop to 6.25% of the original concentration.
Explanation:
Integrated rate law of first order kinetic is given as:
[tex]\log [A]=\log [A_o]-\frac{kt}{2.303}[/tex]
Where:
[tex][A_o][/tex]= Initial concentration of reactant
[A] =concentration left after time t .
k = Rate constant
We are given :
[tex]k=9.60\times 10^{-3} s^{-1}[/tex]
[tex][A_o]=x[/tex]
[tex][A]=6.25% of x =0.0625x[/tex]
Time taken during the reaction = t=?
[tex]\log [0.0625 x]=\log[x]-\frac{9.60\times 10^{-3} s^{-1}\tmes t}{2.303}[/tex]
t = 288.86 seconds = 4.81 minutes
1 min = 60 seconds
In 4.81 minutes the concentration of reactant will drop to 6.25% of the original concentration.
