# For a first order reaction if the time for 50%, 75% and 87.5% changes are t1, t2 and t3 respectively then what would be the value of t1: t2: t3 ?

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Mar 9, 2018

${t}_{1} : {t}_{2} : {t}_{3} = 1 : 2 : 3$

#### Explanation:

Consider a first-order reaction

$\text{A →Products}$

The time required for the initial concentration ${\left[\text{A}\right]}_{0}$ to drop to

• $\frac{1}{2} {\left[\text{A}\right]}_{0}$ is one half-life
• $\frac{1}{4} {\left[\text{A}\right]}_{0}$ is two half-lives
• $\frac{1}{8} {\left[\text{A}\right]}_{0}$ is three half-lives

Also,

t_½ = ln2/k. so each half-life takes the same amount of time.

The ratios are

${t}_{1} : {t}_{2} : {t}_{3} = \text{1 half-life:2 half-lives:3 half-lives = 1:2:3}$

Here is a typical plot for a first-order reaction.

The initial concentration is 1.7.

It takes

• 400 s for the concentration to drop by 50 % (to 0.85)
• 800 s for the concentration to drop by 75 % (0.42)
• 400 s for the concentration to drop by 87.5 % (to 0.21)

The times are in the ratio $\text{1:2:3}$.

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Al E. Share
Mar 8, 2018

Recall, for a first order elementary reaction,

$\ln {\left[A\right]}_{\text{t}} = - k t + \ln {\left[A\right]}_{0}$

$\implies \ln \left(\frac{{\left[A\right]}_{\text{t}}}{{\left[A\right]}_{0}}\right) = - k t$

Without more data, I couldn't give you a real value, but I could give you one as a factor of $k$ (the rate constant).

We can conceptually assume that ${\left[A\right]}_{0} = 1 M$ to derive our data.

Hence,

$\ln \left(\frac{{\left[A\right]}_{\text{t}}}{{\left[A\right]}_{0}}\right) = - k t$

$\implies {t}_{1} = - \ln \frac{0.5}{k} \approx \frac{0.693}{k}$

${t}_{2} = - \ln \frac{0.75}{k} \approx \frac{0.288}{k}$, and

${t}_{3} = - \ln \frac{0.875}{k} \approx \frac{0.134}{k}$

Moreover,

${t}_{1} : {t}_{2} : {t}_{3} \approx 5.17 : 2.15 : 1$

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