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).

  • $\begingroup$ 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.
    Commented 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
    Commented Jul 25, 2023 at 11:51
  • $\begingroup$ Start with $dC/dt =-k$ integrate $C_0$ to $C$ and then make the half life. Take logs of this and look at the slope of the plot you get with that given in your question and you should get eqn(1) which has a slope of 1. $\endgroup$
    – porphyrin
    Commented Jul 19 at 9:13

3 Answers 3


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 $0.5$).

  • 1
    $\begingroup$ 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.
    Commented Mar 3, 2022 at 0:17

The generic slope-intercept equation for a line is:


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


In this case:


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



We can see that:





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.


For an $n$-th order reaction, $$ -\frac{\text{d}a}{\text{d}t}=ka^n $$ Integrating between the limit $0$ and $t_{1/2}$

$$\begin{align} k\int_0^{t_{1/2}}{\text{d}t} &= -\int_a^{a/2}{\frac{1}{a^n}\ \text{d}a} \end{align}$$

$$\begin{equation}\begin{aligned} kt_{1/2} &=-\frac{1}{1-n}\left(a^{1-n}\right)_a^{a/2} \\ &= \frac{1}{n-1}\left[\left(\frac{a}{2}\right)^{1-n}-a^{1-n}\right] \\ &= \frac{2^{n-1}-1}{n-1}a^{1-n} \\ t_{1/2} &= \frac{1}{k}\frac{2^{n-1}-1}{n-1}a^{1-n} \end{aligned}\end{equation}$$

$$ \ln(t_{1/2})=\ln\left(\frac{1}{k}\frac{2^{n-1}-1}{n-1}\right)+(1-n)\ln(a) $$ The slope of the straight line equation is $m=1-n$. Comapring with the graph, where $m=1$, we can see that $n=0$. The intercept, for $n=0$, is $-\ln(2k)$, and since in the graph slope is positive, what we can conclude is that $\bbox[yellow] {k<\ce{0.5 mol\ \ell^{-1}\ s^{-1}}} $


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