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Consider the reaction [tex]\(aA \rightarrow \text{Products}\)[/tex].

Given: [tex]\([A]_0 = 5.0 \, M\)[/tex] and [tex]\(k = 1.0 \times 10^{-2}\)[/tex].

Calculate [tex]\([A]\)[/tex] after 30.0 seconds for each case:

a) Zero order
i. Concentration [tex]\([A]\)[/tex]
ii. Half-life

b) First order
i. Concentration [tex]\([A]\)[/tex]
ii. Half-life

c) Second order
i. Concentration [tex]\([A]\)[/tex]
ii. Half-life

Sagot :

GinnyAnswer
Let's start by breaking down the problem and solving each part step-by-step.

Given:

- Initial concentration, [tex]\( [A]_0 = 5.0 \, \text{M} \)[/tex]
- Rate constant, [tex]\( k = 1.0 \times 10^{-2} \)[/tex]
- Time elapsed, [tex]\( t = 30.0 \, \text{s} \)[/tex]

### a) Zero Order Reaction

For a zero order reaction, the rate law is given by:
[tex]\[ [A] = [A]_0 - kt \][/tex]

i. To find the concentration of [tex]\( A \)[/tex] after 30.0 seconds:
[tex]\[ [A] = 5.0 \, \text{M} - (1.0 \times 10^{-2}) \times 30.0 \][/tex]
[tex]\[ [A] = 5.0 \, \text{M} - 0.3 \, \text{M} \][/tex]
[tex]\[ [A] = 4.7 \, \text{M} \][/tex]

ii. The half-life for a zero order reaction is:
[tex]\[ t_{1/2} = \frac{[A]_0}{2k} \][/tex]
[tex]\[ t_{1/2} = \frac{5.0 \, \text{M}}{2 \times 1.0 \times 10^{-2}} \][/tex]
[tex]\[ t_{1/2} = \frac{5.0}{0.02} \][/tex]
[tex]\[ t_{1/2} = 250.0 \, \text{s} \][/tex]

### b) First Order Reaction

For a first order reaction, the rate law is given by:
[tex]\[ [A] = [A]_0 e^{-kt} \][/tex]

i. To find the concentration of [tex]\( A \)[/tex] after 30.0 seconds:
[tex]\[ [A] = 5.0 \, \text{M} \times e^{-(1.0 \times 10^{-2}) \times 30.0} \][/tex]
[tex]\[ [A] = 5.0 \, \text{M} \times e^{-0.3} \][/tex]
[tex]\[ [A] \approx 5.0 \, \text{M} \times 0.7408182 \][/tex]
[tex]\[ [A] \approx 3.704 \, \text{M} \][/tex]

ii. The half-life for a first order reaction is:
[tex]\[ t_{1/2} = \frac{\ln 2}{k} \][/tex]
[tex]\[ t_{1/2} = \frac{0.693}{1.0 \times 10^{-2}} \][/tex]
[tex]\[ t_{1/2} \approx 69.3147 \, \text{s} \][/tex]

### c) Second Order Reaction

For a second order reaction, the rate law is given by:
[tex]\[ \frac{1}{[A]} = \frac{1}{[A]_0} + kt \][/tex]

i. To find the concentration of [tex]\( A \)[/tex] after 30.0 seconds:
[tex]\[ \frac{1}{[A]} = \frac{1}{5.0 \, \text{M}} + (1.0 \times 10^{-2}) \times 30.0 \][/tex]
[tex]\[ \frac{1}{[A]} = 0.2 + 0.3 \][/tex]
[tex]\[ \frac{1}{[A]} = 0.5 \][/tex]
[tex]\[ [A] = \frac{1}{0.5} \][/tex]
[tex]\[ [A] = 2.0 \, \text{M} \][/tex]

ii. The half-life for a second order reaction is:
[tex]\[ t_{1/2} = \frac{1}{k[A]_0} \][/tex]
[tex]\[ t_{1/2} = \frac{1}{(1.0 \times 10^{-2}) \times 5.0} \][/tex]
[tex]\[ t_{1/2} = \frac{1}{0.05} \][/tex]
[tex]\[ t_{1/2} = 20.0 \, \text{s} \][/tex]

In summary:

- a) Zero order:
- Concentration after 30s: [tex]\( [A] = 4.7 \, \text{M} \)[/tex]
- Half-life: [tex]\( t_{1/2} = 250.0 \, \text{s} \)[/tex]

- b) First order:
- Concentration after 30s: [tex]\( [A] \approx 3.704 \, \text{M} \)[/tex]
- Half-life: [tex]\( t_{1/2} \approx 69.3147 \, \text{s} \)[/tex]

- c) Second order:
- Concentration after 30s: [tex]\( [A] = 2.0 \, \text{M} \)[/tex]
- Half-life: [tex]\( t_{1/2} = 20.0 \, \text{s} \)[/tex]
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