Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • At <img alt="$$ \begin{aligned} & \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\

At 298 \mathrm{~K}
$$ \begin{aligned} & \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}), \mathrm{K}_1=4 \times 10^5 \\ & \mathrm{~N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}), \mathrm{K}_2=1.6 \times 10^{12} \\ & \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(\mathrm{g}), \mathrm{K}_3=1.0 \times 10^{-13} \end{aligned}
Based on above equilibria, the equilibrium constant of the reaction,

2 \mathrm{NH}_3(\mathrm{~g})+\frac{5}{2} \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) is  ___________ \times 10^{-33} (Nearest integer).

Option: 1

4


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Reverse equation (1)So \, \, \mathrm{K}_1^1=\frac{1}{\mathrm{~K}_1}\, \, \, \, \, \, \, \, \, \, \,... (a)
$$ +
Add equation   ......(2) \, \, \mathrm{K}_1^1=\mathrm{K}_2\, \, \, \, \, \, \, \, \, ....(b)

+
Multiply equation (3) by (3)  \mathrm{K}_3^1=\mathrm{k}_3^3\, \, \, \, \, \, \, ....(c)
Add (a), (b) \& (c)
$$ \begin{aligned} & \mathrm{K}_{\mathrm{c}}^1=\frac{\mathrm{K}_2 \times \mathrm{K}_3^3}{\mathrm{~K}_1}=\frac{1.6 \times 10^{12} \times 1 \times 10^{-39}}{4 \times 10^5} \\ & \Rightarrow 4 \times 10^{-33} \end{aligned}

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: BTech CSE cut-offs for PwD candidates at NITs make entry easy; top institutes still selective
anam-khan

73 or 99 percentile? JEE Main result questioned, J-K candidate says, ‘Don’t create hype’
anam-khan

JEE Main 2025: 11 from Kota coaching centres among 24 students with 100 percentile

Latest Question