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For a first-order reaction, the half-life is constant. It depends only on the rate constant and not on the reactant concentration. It is expressed as: t1/2=0.693/k

For a second-order reaction, the half-life depends on the rate constant and the concentration of the reactant and so is expressed as: t1/2= 1/k[A]0
A. A certain first-order reaction (A-->Products) has a rate constant of 3.30×10^-3 s^-1 at 45 degrees C. How many minutes does it take for the concentration of the reactant, [A], to drop to 6.25% of the original concentration?
B. A certain second-order reaction (B-->Products) has a rate constant of 1.70×10^-3 M^-1*s^-1 at 27 degrees C and an initial half-life of 296 s. What is the concentration of the reactant B after one half-life?

Respuesta :

Answer:

A.  14 min

B . 0.995 M

Explanation:

A.

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:

The final concentration is dropped to 6.25 % of the initial concentration. SO,

[tex]\frac {[A_t]}{[A_0]}[/tex] = 0.0625

k = [tex]3.30\times 10^{-3}\ s^{-1}[/tex]

So,

[tex]\frac {[A_t]}{[A_0]}=e^{-k\times t}[/tex]

[tex]0.0625=e^{-3.30\times 10^{-3}\times t}[/tex]

[tex]\ln \left(0.0625\right)=\ln \left(e^{-3.3\times \:10^{-3}t}\right)[/tex]

[tex]\ln \left(0.0625\right)=-3.3\times \:10^{-3}t[/tex]

[tex]t=840.18\ s[/tex]

Also, 1 s = 1/60 minutes.

So, [tex]t=\frac{840.18}{60}=14\ min[/tex]

14 minutes it takes for the concentration of the reactant to drop to 6.25% of the original concentration.

B.

(a) Half life expression for second order kinetic is:

[tex]t_{1/2}=\frac{1}{k[A_o]}[/tex]

Where,  

[tex][A_o][/tex] is the initial concentration = ?

k is the rate constant = [tex]1.70\times 10^{-3}[/tex] M⁻¹s⁻¹

Half life = 296 s

So,

[tex]296=\frac{1}{1.70\times 10^{-3}\times [A_o]}[/tex]

[tex]296=\frac{1000}{1.7[A_o]}[/tex]

[tex][A_o]=\frac{1250}{629}[/tex]

[tex][A_o]=1.99\ M[/tex]

Half life is the time at which the concentration of the reactant reduced to half. So, concentration remains = [tex]\frac{1.99}{2}\ M[/tex] = 0.995 M

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