Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • 4.0 moles of argon and 5.0 moles of  are introduced into an evacuated flask

4.0 moles of argon and 5.0 moles of \mathrm{PCl}_{5} are introduced into an evacuated flask of 100 litre capacity at 610 K. The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be 6.0 atm. The \mathrm{K}_{\mathrm{p}} for the reaction is 

\left[\text { Given: } \mathrm{R}=0.082 \mathrm{~L} \operatorname{atm~} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right] 

Option: 1

2.25


Option: 2

6.24


Option: 3

12.13


Option: 4

15.24


Answers (1)

best_answer

Total pressure inside the flask

\mathrm{P_{\text {Total }}=\frac{9 \times 0.082 \times 610}{100}=4.5 \mathrm{~atm}} \\

\mathrm{P_{\mathrm{PCl_5 }}=\frac{5}{9} \times 4.5=2.5 \mathrm{~atm}} \\

\mathrm{P_{\text {Ar }}=\frac{4}{9} \times 4.5=2 \mathrm{~atm} }

Now,

\mathrm{P_{\text {Total }} \text { at\: equilibrim } =2.5-p+p+p+p_{Ar}=6} \\

\mathrm{p =1.5} \\

\mathrm{k_{p}=\frac{1.5 \times 1.5}{2.5-1.5} =2.25 }

Hence the correct answer is option 1

Posted by

Rishi

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: Most important chapters for physics, chemistry, mathematics to score good marks
anam-khan

JEE Main 2025 session 2 dates clash with ISC, PSEB 12th exams; no impact on major science subjects
anam-khan

Delhi police EOW books FIITJEE for fraud; 192 complaints filed

Latest Question