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Question
The slope in the plot of `log ["R"]_0/(["R"])` Vs. time for a first-order reaction is ______.
Options
`(+"k")/2.303`
+k
`(-"k")/2.303`
−k
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Solution
The slope in the plot of `log ["R"]_0/(["R"])` Vs. time for a first-order reaction is `bbunderline((+"k")/2.303)`.
Explanation:
For the reaction, \[\ce{R -> Products}\]
The integrated first-order rate equation is
log [R] = `(−"kt")/2.303 + log ["R"_0]`
or `log (["R"_0])/(["R"]) = ("kt")/2.303`
Comparing the above equation with a straight-line equation y = mx + c
∴ y = `log (["R"_0])/(["R"])`, x = t, slope = `"k"/2.303`
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