For the graphical representation of the first-order reaction

The slope of the straight line is

\(\frac{k}{2.303}\)

The answer is (3) \(\frac{k}{2.303}\).

We know for a first-order reaction,

\(k = \frac{2.303}{t}log\frac{[R_0]}{[R]}\)

or, \(\frac{k}{2.303}t = log\frac{[R_0]}{R} = log[R_0] - log[R]\)

or, \(log[R] = -\frac{k}{2.303}t + log[R_0]\)

This is the equation of the straight line \((y = mx + c)\). Thus, if \(log[R]\) or \(log(a - x)\) values are plotted against time ‘\(t\)’, the graph obtained should be a straight line if the reaction is of the first order.

The intercept made on the y-axis would be ‘\(log[R_0]\)’ and the slope of the line would be equal to \(\left(-\frac{k}{2.303}\right)\), i.e., \(\text{slope = }-\frac{k}{2.303}\).

Further, we can also write

\(log\frac{[R_0]}{[R]} = \frac{k}{2.303}t\)

Thus if we plot \( log\frac{[R_0]}{R}\) versus \(t\), a straight line graph will be obtained with \(\text{slope = }\frac{k}{2.303}\)