At the point of intersection of the two curves shown, the concentration of textB is given by, for the first reaction textA arrow nB.

Question

At the point of intersection of the two curves shown, the concentration of textB is given by, for the first reaction textA arrow nB. <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/c-mmgdcmeiaxki.png" /> (A) ( textn textA0/2) (B) ( textA0/ textn - 1) (C) ( textn textA0/ textn + 1) (D) ((n-1/n+1)) A0

Tardigrade Admin · Accepted Answer

According to question, at presentation point ⇒ [ textA] = [ textB] ⇒ A0-x=n x ⇒ A0=(n+1) x ⇒ x=(A0/n+1) [B]=n x =((n × A0/n+1))

Q. At the point of intersection of the two curves shown, the concentration of $B$ is given by________, for the first reaction $A \to n B .$

$\frac{n A _{0}}{2}$

$\frac{A _{0}}{n - 1}$

$\frac{n A _{0}}{n + 1}$

$(\frac{n - 1}{n + 1}) A_{0}$

According to question, at presentation point $\Rightarrow [A] = [B]$

$\Rightarrow A_{0} - x = n x$
$\Rightarrow A_{0} = (n + 1) x$
$\Rightarrow x = \frac{A _{0}}{n + 1}$
$[B] = n x$
$= (\frac{n \times A _{0}}{n + 1})$

Q. At the point of intersection of the two curves shown, the concentration of B is given by________, for the first reaction A→nB.

2nA0​​

n - 1A0​​

n+1nA0​​

(n+1n−1​)A0​