The equilibrium constant for the given reacti | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For the reaction $N_{2(g)} + 2 O_{2(g)} \rightleftharpoons 2 NO_{2(g)}$,the equilibrium constant is $K_1 = 100$.The expression is $K_1 = \frac{[NO

Vedclass · Accepted Answer

$0.1$

Vedclass · Answer

(C) For the reaction $N_{2(g)} + 2 O_{2(g)} \rightleftharpoons 2 NO_{2(g)}$,the equilibrium constant is $K_1 = 100$.The expression is $K_1 = \frac{[NO_2]^2}{[N_2][O_2]^2} = 100$.For the reaction $NO_{2(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + O_{2(g)}$,the equilibrium constant is $K_2 = \frac{[N_2]^{1/2} [O_2]}{[NO_2]}$.Comparing $K_2$ with $K_1$,we see that $K_2 = \sqrt{\frac{1}{K_1}}$.Therefore,$K_2 = \sqrt{\frac{1}{100}} = \frac{1}{10} = 0.1$.

The equilibrium constant for the given reaction is $100$.
$N_{2(g)} + 2 O_{2(g)} \rightleftharpoons 2 NO_{2(g)}$
What is the equilibrium constant for the reaction given below?
$NO_{2(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + O_{2(g)}$

Similar Questions

The set of all points for which $f(x) = x^2 \cdot e^{-x}$ strictly increases is

The molar mass of a solute $X$ in $g \ mol^{-1}$,if its $1 \%$ solution is isotonic with a $5 \%$ solution of cane sugar (molar mass $= 342 \ g \ mol^{-1}$),is

If $5 \cos x + 12 \cos y = 13$,then the maximum value of $5 \sin x + 12 \sin y$ is:

In ruby laser,the stimulated emission is due to transition from

Which of the following represents a correct group of Embryophyta?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The equilibrium constant for the given reaction is $100$.$N_{2(g)} + 2 O_{2(g)} \rightleftharpoons 2 NO_{2(g)}$What is the equilibrium constant for the reaction given below?$NO_{2(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + O_{2(g)}$