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For integrated rate laws, I have attempted to find the shelf-life and half-life of second-order reaction:


handwritten work


However, the answers I have obtained (which are labelled with a red cross) are very different from the correct answers (bottom-right corner). I am wondering which parts were done incorrectly.

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    $\begingroup$ For one thing, how did you obtain negative values? As you're solving the problem, that should immediately cause you to think that something is amiss. $\endgroup$
    – Zhe
    Commented Jun 7, 2020 at 16:38
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    $\begingroup$ Also, when you're solving for $t_{10\%}$, you need to substitute $[\ce{A}]=0.9[\ce{A}]_{0}$. The power of $1-n$ also applies to the 0.9. $\endgroup$
    – Zhe
    Commented Jun 7, 2020 at 16:40
  • $\begingroup$ Thank you so much Zhe! I did realise that I obtained a negative value (which is horribly wrong). I just could not figure out where went wrong. You were exactly right on my mistake - 0.9 applies to the exponent of 1-n as well. $\endgroup$
    – Bohan
    Commented Jun 7, 2020 at 16:44
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    $\begingroup$ Posting image of hand written text or formulas is highly discouraged, as it puts burden of reading, evaluating, referring and reusing on shoulders of responders. Furthermore, it cannot be indexed and searched for. Consider rewriting it. $\endgroup$
    – Poutnik
    Commented Jun 7, 2020 at 16:48
  • $\begingroup$ Thank you for the critique @Poutnik - I was being quite lazy (and struggling to learn the formula function as well). $\endgroup$
    – Bohan
    Commented Jun 7, 2020 at 17:00

1 Answer 1

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Your mathematics is correct until you calculated for specific cases. Let's go back to your $n$th order version:

$$\frac{[\ce{A}]^{1-n}}{1-n} = -k_nt + \frac{[\ce{A}]_\circ^{1-n}}{1-n} \ \text{where } n \ne 1 \tag1$$

At this point, since you are working on second order kinetic, it is easy if you substitute $n = 2$ in equation $(1)$:

$$\frac{[\ce{A}]^{-1}}{-1} = -k_2t + \frac{[\ce{A}]_\circ^{-1}}{-1} $$

Once simplify, it becomes:

$$\frac{1}{[\ce{A}]} = k_2t + \frac{1}{[\ce{A}]_\circ} \tag2$$

For self life, substitute $t = t_{10\%}$ and ${[\ce{A}]} = 0.9{[\ce{A}]_\circ}$ in equation $(2)$:

$$\frac{1}{0.9[\ce{A}]_\circ} = k_2t_{10\%} + \frac{1}{[\ce{A}]_\circ} $$

$$\therefore \; t_{10\%} = \frac{1}{k_2}\left(\frac{1}{0.9[\ce{A}]_\circ} - \frac{1}{[\ce{A}]_\circ} \right) = \frac{1}{k_2[\ce{A}]_\circ}\left(\frac{1}{0.9} - 1 \right) = \frac{1}{k_2[\ce{A}]_\circ}\left(\frac{1}{9} \right) = \bbox[yellow]{\frac{0.11}{k_2[\ce{A}]_\circ}}$$

Similarly, for half life, substitute $t = t_{50\%}$ and ${[\ce{A}]} = 0.5{[\ce{A}]_\circ}$ in equation $(2)$:

$$\frac{1}{0.5[\ce{A}]_\circ} = k_2t_{50\%} + \frac{1}{[\ce{A}]_\circ} $$

$$\therefore \; t_{50\%} = \frac{1}{k_2}\left(\frac{1}{0.5[\ce{A}]_\circ} - \frac{1}{[\ce{A}]_\circ} \right) = \frac{1}{k_2[\ce{A}]_\circ}\left(\frac{1}{0.5} - 1 \right) = \frac{1}{k_2[\ce{A}]_\circ}\left(\frac{0.5}{0.5} \right) = \bbox[yellow]{\frac{1}{k_2[\ce{A}]_\circ}}$$

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  • $\begingroup$ @Rodrigo de Azevedo: You can find the definition for standard state in any chemistry textbook, by the way. $\endgroup$ Commented Jun 28, 2020 at 18:05

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