You visited us 1 times! Enjoying our articles? Unlock Full Access!

Order of Reaction

Aaq⟶ Baq+Caq ...

Question

$A (a q) ⟶ B (a q) + C (a q)$ is a first order reaction.
Time
$t$
$\infty$
moles of reagent
$n_{1}$
$n_{2}$
Reaction progress is measure with help of titration ' $R$ '. If all $A, B$ and $C$ reacted with reagent and have ' $n$ ' factors [ $n$ factor; $e q . w t = \frac{m o l . w t .}{n}$ ] in the ratio of $1 : 2 : 3$ with the reagent. The $k$ in terms of $t, n_{1}$ and $n_{2}$ is :

k = \frac{1}{t} ln (\frac{n_{2}}{n_{2} - n_{1}})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

k = \frac{1}{t} ln (\frac{2 n_{2}}{n_{2} - n_{1}})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

k = \frac{1}{t} ln (\frac{4 n_{2}}{n_{2} - n_{1}})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

k = \frac{1}{t} ln (\frac{4 n_{2}}{5 (n_{2} - n_{1})})

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $k = \frac{1}{t} ln (\frac{4 n_{2}}{5 (n_{2} - n_{1})})$
The expression for the rate constant for the first order reaction is $k = \frac{1}{t} ln (\frac{[A]_{0}}{[A]_{t}})$
But $(\frac{[A]_{0}}{[A]_{t}}) = (\frac{4 n_{2}}{5 (n_{2} - n_{1})})$
The expression for the rate constant for the first order reaction becomes $k = \frac{1}{t} ln (\frac{4 n_{2}}{5 (n_{2} - n_{1})})$

Suggest Corrections

Similar questions

n_{1}, n_{2}, \dots, n_{k}

n

1

n .

n_{1} + n_{2} + \dots + n_{k} = 91,

\frac{1}{n_{1}} + \frac{1}{n_{2}} + \dots + \frac{1}{n_{k}}

t_{n} = \sum_{r = 0}^{n} \frac{1}{{({^{n} C}_{r})}^{k}}

S_{n} = \sum_{r = 0}^{n} \frac{r}{{({^{n} C}_{r})}^{k}}

k \in Z^{+}

{cos}^{- 1} (\frac{S_{n}}{n t_{n}})

ln \frac{K_{2}}{K_{1}} = \frac{{Δ H}^{o}}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]

{Δ H}^{o} = - v e

\frac{K_{2}}{K_{1}} < 1

n_{1}

n_{2}

△ E = R_{H} \times c [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]

A λ \to B 2 λ - \to C

T = 0

N_{0}

0

0

T = T

N_{1}

N_{2}

N_{3}

N_{1}

N_{2}

N_{2}

Join BYJU'S Learning Program