Which one of the following graph is not applicable for a 1st order reaction ($R \rightarrow P$)?

For a first-order reaction $R \rightarrow P$, the integrated rate law is given by:

$[R] = [R]_0 e^{-kt}$

This indicates that the concentration of the reactant $[R]$ decreases exponentially with time $t$. Therefore, the graph of $[R]$ versus $t$ should be an exponentially decaying curve. Graph (A) shows $[R]$ as a constant horizontal line, which is incorrect for a first-order reaction.

Taking the natural logarithm on both sides of the rate law:

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

This is a linear equation ($y = mx + c$) with a negative slope ($-k$). Graph (B) correctly represents this.

Converting to base-10 logarithm:

$\log_{10} [R] = \log_{10} [R]_0 - \dfrac{k}{2.303} t$

This is also a straight line with a negative slope ($-\dfrac{k}{2.303}$). Graph (C) correctly represents this.

Rearranging the above equation:

$\log_{10} \dfrac{[R]_0}{[R]} = \dfrac{k}{2.303} t$

This represents a straight line passing through the origin with a positive slope ($\dfrac{k}{2.303}$). Graph (D) correctly represents this.

Thus, the graph in option (A) is not applicable for a first-order reaction.

Answer: