For an elementary reaction aA- Product- the graph plotted between log -d-A-dt - vs- time gives a straight line with intercept equal to 0-6 and showing an angle of 45oC with origin- then-

The correct option is A rate constant =3.98 time^ 1 and a=1 Rate= dC_Adt=KC_A^a ∴ log [ dC_Adt ]=log K+alog C_A ∴ log K=0.6 ⇒ ...

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Standard XII

Mathematics

Rotation Concept

For an elemen...

Question

For an elementary reaction $a A \to P r o d u c t$ , the graph plotted between $l o g [\frac{- d [A]}{d t}]$ vs. time gives a straight line with intercept equal to 0.6 and showing an angle of $45^{o} C$ with origin, then:

rate constant

= 3.98 t i m e^{- 1}

and

a = 1

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rate constant

= 3.98 m o l L^{- 1} t^{- 1}

and

a = 1

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rate constant

= 1.99 t i m e^{- 1}

and

a = 1

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rate constant

= 1.99 m o l L^{- 1} t^{- 1}

and

a = 2

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Solution

The correct option is A rate constant $= 3.98 t i m e^{- 1}$ and $a = 1$
$R a t e = - \frac{d C_{A}}{d t} = K C_{A}^{a}$

$∴ l o g [- \frac{d C_{A}}{d t}] = l o g K + a l o g C_{A}$

$∴ l o g K = 0.6$
$\Rightarrow K = 3.98 t i m e^{- 1}$ and

$a = 1 = t a n 45^{o}$

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For an elementary reaction aA→Product, the graph plotted between log[−d[A]dt] vs. time gives a straight line with intercept equal to 0.6 and showing an angle of 45oC with origin, then:

The correct option is A rate constant =3.98 time−1 and a=1Rate=−dCAdt=KCaA∴log[−dCAdt]=logK+alogCA∴logK=0.6⇒K=3.98 time−1 and a=1=tan 45o

For an elementary reaction $a A \to P r o d u c t$ , the graph plotted between $l o g [\frac{- d [A]}{d t}]$ vs. time gives a straight line with intercept equal to 0.6 and showing an angle of $45^{o} C$ with origin, then:

The correct option is A rate constant $= 3.98 t i m e^{- 1}$ and $a = 1$
$R a t e = - \frac{d C_{A}}{d t} = K C_{A}^{a}$

$∴ l o g [- \frac{d C_{A}}{d t}] = l o g K + a l o g C_{A}$

$∴ l o g K = 0.6$
$\Rightarrow K = 3.98 t i m e^{- 1}$ and

$a = 1 = t a n 45^{o}$