Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • For the reaction: <img alt="\mathrm{Fe_{2}N(s)+\frac{3}{2}H_{2}(g)\rightleftharpoons 2Fe(s)+NH_{3}(g)}" src="/latex-image/?%5Cmathrm%7BFe_%7B2%7DN%28s%29&plus;%5Cfrac%7B3%7D%7B2%7DH_%7B2%7D%28g%29%5Crightleftharpoons%202Fe%28s%29&plus;NH_%7B3%7

For the reaction: \mathrm{Fe_{2}N(s)+\frac{3}{2}H_{2}(g)\rightleftharpoons 2Fe(s)+NH_{3}(g)}
Option: 1 K_{c}=K_{p}(RT)^{-\frac{1}{2}}
Option: 2 K_{c}=K_{p}(RT)^{\frac{1}{2}}
Option: 3 K_{c}=K_{p}(RT)  
Option: 4 K_{c}=K_{p}(RT)^{\frac{3}{2}}

Answers (1)

best_answer

The reaction occurs as follows:

\mathrm{Fe_{2}N(s)\: +\: \frac{3}{2}H_{2}(g)\: \rightleftharpoons \: 2Fe(s)\: +\: NH_{3}(g)}

\mathrm{\Delta n\: =\: -\frac{1}{2}}

We know that:

\\\mathrm{k_{p}\: =\: k_{c}(RT)^{\Delta n}}\\\\\mathrm{Thus,\: k_{p}\: =\: k_{c}(RT)^{-\frac{1}{2}}}\\\\\mathrm{k_{c}\: =\: k_{p}(RT)^{\frac{1}{2}}}

Therefore, Option(2) is correct.

Posted by

Kuldeep Maurya

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 Registration Live: NTA session 1 application last date today; official website, syllabus
anam-khan

JEE Main 2025 session 1 registration ends today at jeemain.nta.nic.in; application correction from November 26
anam-khan

JEE Main 2025 registration ends tomorrow; changes introduced in exam this year

Latest Question