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Standard XII

Chemistry

Spontaneity

ABC T=0 ...

Question

$A λ \to B 2 λ - \to C$
$T = 0$ $N_{0}$ $0$ $0$
$T = T$ $N_{1}$ $N_{2}$ $N_{3}$
The ratio of $N_{1}$ to $N_{2}$ when $N_{2}$ is maximum is :

At no time this is possible

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2

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1 / 2

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\frac{l n 2}{2}

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Solution

The correct option is A $2$
First order decay:
$\frac{d N_{1}}{d t} = - λ N_{1}$ ----------------(1)
$\frac{d N_{2}}{d t} = λ N_{1} - 2 λ N_{2}$ ----------(2)
$\frac{d N_{3}}{d t} = 2 λ N_{2}$ ---------------(3)
When $N_{2}$ is maximum:
$\frac{d N_{2}}{d t} = 0$
$0 = λ N_{1} - 2 λ N_{2}$
$\frac{N_{1}}{N_{2}} = 2$

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Aλ→B2λ−→CT=0 N0 0 0T=T N1 N2 N3The ratio of N1 to N2 when N2 is maximum is :

The correct option is A 2First order decay:dN1dt=−λN1 ----------------(1)dN2dt=λN1−2λN2 ----------(2)dN3dt=2λN2 ---------------(3)When N2 is maximum:dN2dt=00=λN1−2λN2N1N2=2

$A λ \to B 2 λ - \to C$
$T = 0$ $N_{0}$ $0$ $0$
$T = T$ $N_{1}$ $N_{2}$ $N_{3}$
The ratio of $N_{1}$ to $N_{2}$ when $N_{2}$ is maximum is :