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A graph plotted between $log \: t_{50 \%}$ vs loga concⁿ is a straight line. What conclusion can you drow from the given graph?

(n=order)

Option 1)
$n=1, t_{\frac{1}{2}}=\frac{1}{k.a}$

Option 2)
$n=2, t_{\frac{1}{2}}=\frac{1}{a}$

Option 3)
$n=1, t_{\frac{1}{2}}=\frac{0.693}{k}$

Option 4)
none

Answers (1)

best_answer

As we leaned in concept

First Order Reaction -

The rate of the reaction is proportional to the first power of the concentration of the reaction

- wherein

Formula:

R $\rightarrow$ P

a 0

a-x x

$rate[r]=K[R]^{1}$

$\frac{-d(a-x)}{dt}=K(a-x)$

$\frac{-dx}{dt}=K(a-x)$ [differentiate rate law]

$ln \:[\frac{a}{a-x}]=kt \:(Integrated rate law)$

Unit of $k=sec^{-1}$

$t_\frac{1}{2}=\frac{0.693}{k}$

We know that for first order reactions, $t_{\frac{1}{2}}= \frac{0.693}{k}$

$\log t_{\frac{1}{2}}= \log (0.0693)- \log k$

So, in other words, in a first oder reaction, $t_{\frac{1}{2}}$ does not depend on the concentration.

Therefore, graphically we can draw.

n=1, $t_{\frac{1}{2}}= \frac{0.693}{k}$

Option 1)

$n=1, t_{\frac{1}{2}}=\frac{1}{k.a}$

Incorrect option

Option 2)

$n=2, t_{\frac{1}{2}}=\frac{1}{a}$

Incorrect option

Option 3)

$n=1, t_{\frac{1}{2}}=\frac{0.693}{k}$

Correct option

Option 4)

none

Incorrect option

Posted by

divya.saini

View full answer

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