KCET 2026 Chemistry Solved Paper

Concept:For a first-order reaction, the integrated rate law decides how the concentration and its logarithm change with time.Explanation:For $R \rightarrow P$, the integrated rate law is:$$[R]=[R]_0 e^{-kt}$$So, $[R]$ decreases exponentially with time.A graph showing constant $[R]$ versus $t$ cannot be correct for this reaction.Taking natural logarithms:$$\ln[R]=\ln[R]_0-kt$$This gives a straight line with negative slope $-k$.Hence, graph B is correct.Using common logarithms:$$\log_{10}[R]=\log_{10}[R]_0-\frac{k}{2.303}t$$This also gives a straight line with negative slope.Hence, graph C is correct.Rearranging the same equation:$$\log_{10}\frac{[R]_0}{[R]}=\frac{k}{2.303}t$$This gives a straight line through the origin with positive slope.Hence, graph D is correct.Only graph A, which shows constant $[R]$, is not applicable.Answer:Option A