15:I[8216,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"BackButton"] 16:I[1553,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"QuestionViewer"] 17:T635,

To understand the slope of the plot of $$\log$$ (reactant concentration) against time for a first order reaction, let's start by examining the fundamental relationship governing first order reactions:

A first order reaction can be described by the differential rate law:

$$ \frac{d[A]}{dt} = -k[A] $$

where $$[A]$$ is the concentration of the reactant and $$k$$ is the rate constant.

To solve this differential equation, we can separate variables and integrate:

$$ \int \frac{1}{[A]} \, d[A] = -k \int dt $$

This gives us the integrated form of the first order rate law:

$$ \ln [A] = -kt + C $$

where $$C$$ is the integration constant that can be determined from the initial conditions. If at $$t = 0$$, $$[A] = [A]_0$$ (initial concentration of $$A$$), then:

$$ \ln [A]_0 = C $$

Thus, we can rewrite the equation as:

$$ \ln [A] = \ln [A]_0 - kt $$

The next step is to convert the natural logarithm to a common logarithm (base 10) using the relationship:

$$ \log [A] = \frac{\ln [A]}{\ln 10} $$

This results in:

$$ \log [A] = \log [A]_0 - \frac{kt}{2.303} $$

We can clearly see that this equation is in the form $$y = mx + c$$ where $$y$$ is $$\log [A]$$, $$c$$ is $$\log [A]_0$$, and the slope $$m$$ is $$-k/2.303$$.

Therefore, the slope of the plot of $$\log$$ (reactant concentration) against time for a first order reaction is:<

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