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What will be the order of the reaction for a chemical change having $\log t_{1/2}$ VS $\log a$ Where $a=$ Initial concentration of reactant and $t_{1/2} =$ Half Life?

enter image description here

  1. 0 order
  2. 1st order
  3. 2nd order
  4. None of These

Actually the answer I found by searching in internet is option 1 but following my calculations, I am getting the answer as option 4.

My calculations are as follows:

For 0 order reaction: $$t_{1/2} = \frac{a}{2K}$$

Taking $\log$ on both sides:

$$\log t_{1/2} = \log{\frac{a}{2K}}$$

$$\log t_{1/2} = \log a - \log 2K \tag1$$

This is a Straight line equation of type $y= mx - C $.

MY DOUBT IS If you notice in graph you will get to know that the intercept is positive while in the equation $(1)$, intercept is negative so how is it possible?

My Background- actually I didn't have kept Mathematics as my major subject in senior secondary high school so I have little knowledge with respect to graph so please forgive me if I am wrong at formulating (1).

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  • $\begingroup$ Whoever indicated the precise angle of 45 degrees in that graph should have known that it can only be true for a particular combination of time units (for $t_{1/2}$) and for concentration for ($a$). $\endgroup$
    – Curt F.
    Mar 3, 2022 at 0:20
  • $\begingroup$ At least in the image you have shown, the location of zero of the y-axis is not indicated. How do you know the intercept is positive? $\endgroup$
    – Andrew
    Jul 25, 2023 at 11:51

2 Answers 2

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The equation that you have derived is correct. For a zeroth-order reaction, $$\log t_{1/2}=\log a-\log2K$$ By superficially observing, this seems to be giving a straight line with the equation $y=mx-c$ but actually there are several(not all) reactions in which the $K$ value is very small. In case of any reaction having $K\lt0.5$, $2K$ becomes less than $1$ and hence $\log2K$ becomes negative. Then you would obtain an equation of the form $y=mx+c$ whose graph will have a positive ordinate and it will appear similar to the graph in the question. So that graph in your question is correct for a zero-order reaction having a small rate constant (less than $0.5$).

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    $\begingroup$ There are a lot of units problems/assumptions implicit in your answer. For example, you could just change the time units you use to measure the rate constant. If it has an inconvenient value like 3 mol per L per s, you can just decide to use a time unit of nanoseconds, so then $k$ will be $3\times10^{-9}$ mol per L per nanosecond. And that is much less than 0.5. $\endgroup$
    – Curt F.
    Mar 3, 2022 at 0:17
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The generic slope-intercept equation for a line is:

$$y=mx+c$$

When given angle $\theta$ for a straight line, the slope $m$ is calculated by:

$$m=tan\;\theta$$

In this case:

$$m=tan\;(45°)=1$$

Substituting $m$ into the generic slope-intercept equation for a line, and comparing it with your derived equation (1):

$$y=x+c$$

$$logt_{1/2}=log\;a-log\;(2K)$$

We can see that:

$$y=log\;t_{1/2}$$

$$m=1$$

$$x=log\;a$$

$$c=-log\;(2K)$$

The slope $m$ is not negative, since it's equal to 1.

As far as the intercept $c$ is concerned, as explained by other answers, the value of $K$ for this particular reaction rate is such that taking the base 10 logarithm of $2K$ value will itself result in a negative number, which will then become a positive number once mutiplied by the $-$ sign next to the logarithm function.

In other words, $log(2K)$ is negative and $-log(2K)$ is positive.

In conclusion, both $m$ and $c$ are positive in this problem.

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