N_{2} + 3H_{2}\rightarrow2NH_{3} After completion of 75% of the reaction what would be the mole fraction for the constituents

Answers (1)

As we learnt in

Mole Fraction -

It is the ratio of moles of solute or moles of solvent to moles of the solution.

- wherein

If a substance ‘A’ dissolves in substance ‘B’ and their number of moles are n_{A} and n_{B} respectively; then the mole fractions of A and B are given as

Mole fraction of A = (number of moles of A)/(number of moles of solution ) = n_{A}/(n_{A}+ n_{B})

 

 At   t = \frac{3}{4},

                         N_{2} + 3H_{2}\rightarrow2NH_{3}

t = \frac{3}{4}               \frac{1}{4}            \frac{3}{4}                \frac{3}{2}

Moles of N2 = 0.25

Moles of H2 = 0.75

Moles of NH3 = 1.5

Mole fraction of  N_{2}=\frac{0.25}{2.5} =0.1

Mole fraction of  H_{2}=\frac{0.75}{2.5} =0.3

Mole fraction of  NH_{3}=\frac{1.5}{2.5} =0.6

Most Viewed Questions

Preparation Products

JEE Main Rank Booster 2022

Booster and Kadha Video Lectures, Unlimited Full Mock Test, Adaptive Time Table, 24x7 Doubt Chat Support,.

₹ 6999/- ₹ 799/-
Buy Now
Knockout JEE Main 2022 (One Month)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Live Classes, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 13999/- ₹ 6999/-
Buy Now
Knockout JEE Main 2023 (One Month)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Live Classes, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 13999/- ₹ 6999/-
Buy Now
Knockout JEE Main (One Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Expert Mentorship, - Study Improvement Plan.

₹ 2999/- ₹ 1999/-
Buy Now
Knockout JEE Main (Twelve Months Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Expert Mentorship, - Study Improvement Plan.

₹ 23999/- ₹ 14999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
 
Exams
Articles
Questions