For the reaction,N_2 + O_2 rightleftharpoons | vedclass

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

Question

(C) Let the initial moles of $N_2$ and $O_2$ be $1$ mole each.Let $\alpha$ be the degree of dissociation for both $N_2$ and $O_2$.The reaction is $N_2

Vedclass · Accepted Answer

Both are $\frac{1}{1 + \sqrt{2}}$

Vedclass · Answer

(C) Let the initial moles of $N_2$ and $O_2$ be $1$ mole each.Let $\alpha$ be the degree of dissociation for both $N_2$ and $O_2$.The reaction is $N_2 + O_2 \rightleftharpoons 2NO$.At equilibrium,moles are: $N_2 = (1 - \alpha)$,$O_2 = (1 - \alpha)$,$NO = 2\alpha$.The equilibrium constant $K_c$ is given by: $K_c = \frac{[NO]^2}{[N_2][O_2]} = \frac{(2\alpha)^2}{(1 - \alpha)(1 - \alpha)} = \frac{(2\alpha)^2}{(1 - \alpha)^2}$.Given $K_c = 2$,we have $2 = \left(\frac{2\alpha}{1 - \alpha}\right)^2$.Taking the square root on both sides: $\sqrt{2} = \frac{2\alpha}{1 - \alpha}$.$\sqrt{2}(1 - \alpha) = 2\alpha$.$\sqrt{2} - \alpha\sqrt{2} = 2\alpha$.$\sqrt{2} = \alpha(2 + \sqrt{2})$.$\alpha = \frac{\sqrt{2}}{2 + \sqrt{2}} = \frac{\sqrt{2}}{\sqrt{2}(\sqrt{2} + 1)} = \frac{1}{1 + \sqrt{2}}$.

For the reaction,$N_2 + O_2 \rightleftharpoons 2NO$,the equilibrium constant is $K_c = 2$. If both reactants have the same initial moles,what is the degree of dissociation of $N_2$ and $O_2$?

Similar Questions

At temperature $T$,the dissociation equilibrium of $AB_{2(g)}$ is given as: $2AB_{2(g)} \rightleftharpoons 2AB_{(g)} + B_{2(g)}$. If the degree of dissociation $x$ is very small compared to $1$,then the expression for the equilibrium constant $K_p$ in terms of $x$ and total pressure $P$ is:

If the degree of dissociation for the reaction,$PCl_5 \rightleftharpoons PCl_3 + Cl_2$ is $20\%$ at $1 \ atm$ pressure,calculate $K_c$.

For the reaction $A \rightleftharpoons B + C$,if we start with $2 \ mol$ of $A$ and $0.5 \ mol$ dissociates,what is the degree of dissociation of $A$?

In a Victor Meyer's experiment,$0.23 \ g$ of a volatile solute displaces $112 \ mL$ of air at $NTP$. What is the vapor density of the substance?

Phosphorus pentachloride dissociates as follows,in a closed reaction vessel:
$PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$
If total pressure at equilibrium of the reaction mixture is $P$ and degree of dissociation of $PCl_5$ is $x,$ the partial pressure of $PCl_3$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module