A graph is plotted between $log N v s$ . time for the first-order reaction. Slope and intercept of the graph are :

- \frac{λ}{2.303}, log N_{0}

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\frac{λ}{2.303}, log N_{0}

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- \frac{λ}{2.303}, N_{0}

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\frac{λ}{2.303}, N_{0}

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Solution

The correct option is A $- \frac{λ}{2.303}, log N_{0}$
As we know, $N = N_{o} e^{- λ t}$
By taking log,

$l o g N = - \frac{λ t}{2.303} + log N_{o}$
So, by straight-line equation, $y = m x + c$
$S l o p e = - \frac{λ}{2.303}$
$I n t e r c e p t = log N_{o}$

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A graph is plotted between logNvs. time for the first-order reaction. Slope and intercept of the graph are :

The correct option is A −λ2.303,logN0As we know, N=Noe−λtBy taking log, logN=−λt2.303+logNoSo, by straight-line equation, y=mx+cSlope=−λ2.303Intercept=logNo

A graph is plotted between $log N v s$ . time for the first-order reaction. Slope and intercept of the graph are :

The correct option is A $- \frac{λ}{2.303}, log N_{0}$
As we know, $N = N_{o} e^{- λ t}$
By taking log,

$l o g N = - \frac{λ t}{2.303} + log N_{o}$
So, by straight-line equation, $y = m x + c$
$S l o p e = - \frac{λ}{2.303}$
$I n t e r c e p t = log N_{o}$