SATHEE: Chemical and Ionic Equilibrium

Chemical and Ionic Equilibrium - Result Question 52

####52. The equilibrium constant of the reaction $A_{2} (g) + B_{2} (g) ⇌ 2 A B (g)$ at $100^{\circ} C$ is 50 . If a one litre flask containing one mole of $A_{2}$ is connected to a two litre flask containing two moles of $B_{2}$ , how many moles of $A B$ will be formed at $373 K$ ?

$(1985, 4 M)$

Show Answer

Solution:

$A_{2} (g) + B_{2} (g) ⇌ 2 A B (g)$

$\begin{array}{rc} K = \frac{[A B]^{2}}{[A_{2}] [B_{2}]} = \frac{{(n_{A B})}^{2}}{n_{A_{2}} \cdot n_{B_{2}}} = \frac{(2 x)^{2}}{(1 - x) (2 - x)} \Rightarrow & 50 = \frac{4 x^{2}}{x^{2} - 3 x + 2} \Rightarrow 23 x^{2} - 75 x + 50 = 0 \Rightarrow & x = \frac{75 \pm \sqrt{75^{2} - 4 \times 23 \times 50}}{46} = 0.93, 2.32 \end{array}$

Play Store

App Store

Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on "I understand".

Chemical and Ionic Equilibrium - Result Question 52

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu