For the reaction :NH_{3(g)} rightleftharpoons | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Initial moles: $NH_3 = 1$,$N_2 = 0$,$H_2 = 0$At equilibrium: $NH_3 = (1-\alpha)$,$N_2 = \frac{\alpha}{2}$,$H_2 = \frac{3\alpha}{2}$Total moles at

Vedclass · Accepted Answer

$\alpha = \left( 1 + \frac{3\sqrt{3} P^o}{4 K_P} \right)^{-\frac{1}{2}}$

Vedclass · Answer

(A) Initial moles: $NH_3 = 1$,$N_2 = 0$,$H_2 = 0$At equilibrium: $NH_3 = (1-\alpha)$,$N_2 = \frac{\alpha}{2}$,$H_2 = \frac{3\alpha}{2}$Total moles at equilibrium: $n_{total} = (1-\alpha) + \frac{\alpha}{2} + \frac{3\alpha}{2} = 1 + \alpha$Partial pressures: $p_{NH_3} = \frac{1-\alpha}{1+\alpha} P^o$,$p_{N_2} = \frac{\alpha/2}{1+\alpha} P^o$,$p_{H_2} = \frac{3\alpha/2}{1+\alpha} P^o$$K_P = \frac{(p_{N_2})^{1/2} (p_{H_2})^{3/2}}{p_{NH_3}} = \frac{(\frac{\alpha/2}{1+\alpha} P^o)^{1/2} (\frac{3\alpha/2}{1+\alpha} P^o)^{3/2}}{\frac{1-\alpha}{1+\alpha} P^o}$$K_P = \frac{(\alpha/2)^{1/2} (3\alpha/2)^{3/2}}{(1-\alpha)(1+\alpha)} P^o = \frac{\alpha^{1/2} \cdot 3^{3/2} \cdot \alpha^{3/2}}{2^{1/2} \cdot 2^{3/2} (1-\alpha^2)} P^o = \frac{3\sqrt{3} \alpha^2}{4(1-\alpha^2)} P^o$Rearranging for $\alpha$: $\frac{4 K_P}{3\sqrt{3} P^o} = \frac{\alpha^2}{1-\alpha^2} \implies \frac{3\sqrt{3} P^o}{4 K_P} = \frac{1-\alpha^2}{\alpha^2} = \frac{1}{\alpha^2} - 1$$\frac{1}{\alpha^2} = 1 + \frac{3\sqrt{3} P^o}{4 K_P} \implies \alpha = \left( 1 + \frac{3\sqrt{3} P^o}{4 K_P} \right)^{-1/2}$

For the reaction :
$NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}$ ; $K_P$
The degree of dissociation $(\alpha)$ of $NH_3$ is related to total equilibrium pressure $(P^o)$ as:

Similar Questions

$A$ magnet of length $10 \text{ cm}$ and magnetic moment $1 \text{ Am}^2$ is placed along side $AB$ of an equilateral triangle $ABC$. If the length of the side $AB$ is $10 \text{ cm}$,find the magnetic induction at the point $C$. (Given $\mu_0 = 4\pi \times 10^{-7} \text{ Hm}^{-1}$)

The dimensional formula of $\frac{1}{2} \mu_0 H^2$ (where $\mu_0 = \text{permeability of free space}$ and $H = \text{magnetic field intensity}$) is:

Particles of masses $m, 2m, 3m, \ldots, nm$ grams are placed on the same line at distances $l, 2l, 3l, \ldots, nl$ $cm$ from a fixed point. The distance of the centre of mass of the particles from the fixed point in centimetres is

$A$ line passing through the point of intersection of $x+y=4$ and $x-y=2$ makes an angle $\tan ^{-1}\left(\frac{3}{4}\right)$ with the $x$-axis. It intersects the parabola $y^{2}=4(x-3)$ at points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ respectively. Then, $|x_{1}-x_{2}|$ is equal to

If $0 < y < 2^{1/3}$ and $x(y^3 - 1) = 1$,then $\frac{2}{x} + \frac{2}{3x^3} + \frac{2}{5x^5} + \dots = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module

For the reaction :$NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}$ ; $K_P$The degree of dissociation $(\alpha)$ of $NH_3$ is related to total equilibrium pressure $(P^o)$ as: