# Finding the reaction order from a given plot of chemical kinetics

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? 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).

• 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 ($a$). Mar 3, 2022 at 0:20

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 $$0.5$$).

• 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. Mar 3, 2022 at 0:17

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.