Answer:
After 48.84 s, 18.1% of the reactant will remain.
Explanation:
Given that:
The rate constant, k = 0.0350 s⁻¹
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
[tex][A_0][/tex] is the initial concentration
Given:
18.1 % of the reactant remain which means that 0.181 of [tex][A_0][/tex] is remained. So,
[tex]\frac {[A_t]}{[A_0]}[/tex] = 0.181
t = ?
[tex]\frac {[A_t]}{[A_0]}=e^{-k\times t}[/tex]
[tex]0.181=e^{-0.0350\times t}[/tex]
t = 48.84 s
After 48.84 s, 18.1% of the reactant will remain.
18% of the starting material will remain after 49 seconds.
The term rate of reaction refers to how quickly a reaction proceeds. This reaction as we are told follows first order kinetics.
We can use the formula;
0.693/t1/2 = 2.303/tlog (No/N)
Given that t1/2 = 0.693/k = 0.693/0.0350 s − 1 =19.8 seconds
So;
0.693/19.8 = 2.303/t log 1/0.181
0.693/19.8 = 1.71/t
t = 1.71/0.035
t = 49 seconds
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