Answer:
0.00564 moles
Explanation:
Given that:
The rate constant, k = [tex]6.82\times 10^{-3}[/tex] s⁻¹
Initial concentration [A₀] = [tex]2.90\times 10^{-2}[/tex] mol
Time = 4.0 min = [tex]4.0\times 60[/tex] sec = 240 sec
Using integrated rate law for first order kinetics as:
[tex][A_t]=[A_0]e^{-kt}[/tex]
Where,
[tex][A_t][/tex] is the concentration at time t
So,
[tex][A_t]=2.90\times 10^{-2}\times e^{-6.82\times 10^{-3}\times 240}=2.9\times \frac{1}{10^2}\times \frac{1}{e^{1.6368}}[/tex]
[tex][A_t]=0.00564\ moles[/tex]
The concentration after four minutes is 3.3 ×10−3.
Let us recall that for a first order reaction;
ln[A] = ln[A]o - kt
Where;
[A] = concentration at time t
[A]o = initial concentration
k = rate constant
t = time
[A]o = 2.90×10−2 mol/1.7 L = 0.0171 M
k = 6.82×10−3 s−1
t = 4 min or 240 s
Substituting values;
ln[A] = ln[0.0171 M] - (6.82×10−3 s−1 × 240 s)
[A] = e^ln[0.0171 M] - (6.82×10−3 s−1 × 240 s)
[A] = 3.3 ×10−3.
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