Answer:
Concentration of CH3CHO 5.0 minutes later is
[tex][A] =8.71\times 10^{-3}[/tex]M or 0.00871 M
Explanation:
The concentration for second order reaction can be calculated using the formula:
[tex]\frac{1}{[A]}=\frac{1}{[A_{0}]}+kt[/tex]
Here , [A] = Concentration of substance at time t
t = time
[tex][A_{0}][/tex] = Initial concentration
k = rate constant
According to question :
[A] = Concentration of CH3CHO at time 5.0s = ?
[tex][A_{0}][/tex] = 0.012 M
t = 5.0 min
Convert it into second (because rate constant units are in second)
1 min = 60 s
t = 5 x 60 = 300 s
k = 0.105 M-1 s-1
Insert these value into,
[tex]\frac{1}{[A]}=\frac{1}{[A_{0}]}+kt[/tex]
[tex]\frac{1}{[A]}=\frac{1}{0.012}+ 0.105\times 300[/tex]
[tex]\frac{1}{[A]}=83.33+ 31.5[/tex]
[tex]\frac{1}{[A]}=83.33+ 31.5[/tex]
[tex]\frac{1}{[A]}=114.83[/tex]
[tex][A] =\frac{1}{114.83}[/tex]
[tex][A] =8.71\times 10^{-3}[/tex]M
Hence ,concentration of CH3CHO 5.0 minutes later is
[tex][A] =8.71\times 10^{-3}[/tex]M or 0.00871 M
