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Which of the following graphs is correct for a first-order reaction?

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The answer is the option (i, iv) For a first order reaction

t_{\frac{1}{2}}=\frac{0.693}{k} and rate =-\frac{d[R]}{dt}=k[R]

Since \; k=\frac{2.303}{t}\; log\; [\frac{R^{0}}{R}] the slope for a graph between log log [\frac{R^{0}}{R}] versus t is  \frac{k}{2.303}

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