Question

Draw a plot of ${\text{lo}}{{\text{g}}_{{\text{10}}}}\left[ {\text{R}} \right]$ versus time 't' for a first order reaction. What is the slope of the line equal to?

Answer 1

The plot of ${\text{lo}}{{\text{g}}_{{\text{10}}}}\left[ {\text{R}} \right]$ versus time 't' for a first order reaction is a straight line having negative slope. Write the integrated rate law for the first order reaction as ${\text{lo}}{{\text{g}}_{{\text{10}}}}\left[ {\text{R}} \right]{\text{ }} = {\text{ }}\left( { - \dfrac{k}{{2.303}}} \right)t{\text{ + lo}}{{\text{g}}_{{\text{10}}}}{\left[ {\text{R}} \right]_o}$. Then compare it to the equation of the straight line $Y = mX + C$ .Here, ‘m’ represents the slope.

Complete Step by step answer: Consider the following first order reaction:
${\text{R}} \to {\text{P}}$
Here, R is the reactant and P is the product.
The rate of the first order reaction is given by the formula
$- \dfrac{{d\left[ {\text{R}} \right]}}{{dt}} = \dfrac{{d\left[ {\text{P}} \right]}}{{dt}} = k\left[ {\text{R}} \right]$
Here, $- \dfrac{{d\left[ {\text{R}} \right]}}{{dt}}$ represents the rate at which the reactant R is consumed. It also represents the rate of the reaction. $\dfrac{{d\left[ {\text{P}} \right]}}{{dt}}$ represents the rate of formation of the product P. k is the rate constant or the specific reaction rate. $\left[ {\text{R}} \right]$ represents the concentration of the reactant R.
Write down the integrated rate law for the first order reaction.
${\text{lo}}{{\text{g}}_{{\text{10}}}}\left[ {\text{R}} \right]{\text{ }} = {\text{ lo}}{{\text{g}}_{{\text{10}}}}{\left[ {\text{R}} \right]_o} - \left( {\dfrac{k}{{2.303}}} \right)t$
Rearrange the above rate law expression
${\text{lo}}{{\text{g}}_{{\text{10}}}}\left[ {\text{R}} \right]{\text{ }} = {\text{ }}\left( { - \dfrac{k}{{2.303}}} \right)t{\text{ + lo}}{{\text{g}}_{{\text{10}}}}{\left[ {\text{R}} \right]_o}$
Compare this to the equation of straight line of the type $Y = mX + C$
Here,

$$Y = {\text{lo}}{{\text{g}}_{{\text{10}}}}\left[ {\text{R}} \right] \\\ \Rightarrow m = - \dfrac{k}{{2.303}} \\\ \Rightarrow X = t \\\ \Rightarrow C = {\text{lo}}{{\text{g}}_{{\text{10}}}}{\left[ {\text{R}} \right]_o} \\\$$

Here, m is the slope and C is the Y-intercept.
When you draw a plot of logio[R] versus time 't' for a first order reaction, a straight line with negative slope is obtained.

**The slope of line equal to $- \dfrac{k}{{2.303}}$
The Y-intercept is equal to ${\text{lo}}{{\text{g}}_{{\text{10}}}}{\left[ {\text{R}} \right]_o}$ **

Note: For the first order reaction, the rate of the reaction is directly proportional to the reactant concentration. If you plot ${\text{lo}}{{\text{g}}_{{\text{10}}}}\left[ {\text{R}} \right]$ versus time 't' for the reaction and obtain a straight line with negative slope, then you can say that the reaction is of first order in nature.