The integrated rate equation for a first-order reaction is \(\frac{kt}{2.303} = logR_o – logR_t\). The straight line graph is obtained by plotting:

time vs \(log R_t\)

The correct answer is option 2. time vs \(log R_t\).

We have the integrated rate equation for a first-order reaction:

\(\frac{kt}{2.303} = \log\frac{R_0}{R_t}\)

Where:

\(k\) is the rate constant for the reaction.

\(t\) is time.

\(R_0\) is the initial concentration of the reactant.

\(R_t\) is the concentration of the reactant at time \(t\).

Now, to obtain a straight-line graph, we need to manipulate this equation into the form \(y = mx + c\), where \(y\) is the dependent variable, \(x\) is the independent variable, \(m\) is the slope, and \(b\) is the y-intercept.

First, let's isolate the logarithmic term on one side:
\(\frac{kt}{2.303} = \log R_0 - \log R_t\)

\(log R_t = -\frac{k}{2.303}t + log R_o\)

This equation is comparable to equation of a straight line \((y = mx + c)\).When a graph is plotted between \(log R_t \)and time \(t\) at different intervals of time we get a straight line.