At 600 K,2 mol of NO are mixed with 1 mol | vedclass

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

Question

(C) The reaction is $2 \ NO_{(g)} + O_{2(g)} \rightleftarrows 2 \ NO_{2(g)}$.Initial moles: $NO = 2, O_2 = 1, NO_2 = 0$.At equilibrium,moles of $O_2 =

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) The reaction is $2 \ NO_{(g)} + O_{2(g)} ightleftarrows 2 \ NO_{2(g)}$.Initial moles: $NO = 2, O_2 = 1, NO_2 = 0$.At equilibrium,moles of $O_2 = 1 - x = 0.6$,so $x = 0.4$.Equilibrium moles: $NO = 2 - 2(0.4) = 1.2$,$O_2 = 0.6$,$NO_2 = 2(0.4) = 0.8$.Total moles at equilibrium = $1.2 + 0.6 + 0.8 = 2.6$.Mole fractions: $X_{NO} = \frac{1.2}{2.6}$,$X_{O_2} = \frac{0.6}{2.6}$,$X_{NO_2} = \frac{0.8}{2.6}$.Partial pressures $(P_{total} = 1 \ atm)$: $P_{NO} = \frac{1.2}{2.6} \ atm$,$P_{O_2} = \frac{0.6}{2.6} \ atm$,$P_{NO_2} = \frac{0.8}{2.6} \ atm$.$K_p = \frac{(P_{NO_2})^2}{(P_{NO})^2 	imes P_{O_2}} = \frac{(\frac{0.8}{2.6})^2}{(\frac{1.2}{2.6})^2 	imes (\frac{0.6}{2.6})} = \frac{0.8^2 	imes 2.6}{1.2^2 	imes 0.6} = \frac{0.64 	imes 2.6}{1.44 	imes 0.6} = \frac{1.664}{0.864} \approx 1.926$.The nearest integer is $2$.

Similar Questions

For the reaction $PCl_5 \rightleftharpoons PCl_3 + Cl_2$,the forward reaction at constant temperature is favoured by

At a certain temperature,the equilibrium constant $(K_C)$ is $16$ for the reaction:
$SO_2(g) + NO_2(g) \rightleftharpoons SO_3(g) + NO(g)$
If we take one mole of each of the four gases in a $1 \ L$ container,what would be the equilibrium concentration of $NO$ and $NO_2$ respectively?

At a certain temperature and total pressure of $10^{5} \ Pa$,iodine vapour contains $40 \%$ by volume of $I$ atoms.
$I_{2(g)} \longleftrightarrow 2I_{(g)}$
Calculate $K_{p}$ for the equilibrium.

The standard Gibbs energy change at $300 \, K$ for the reaction $2A \rightleftharpoons B + C$ is $2494.2 \, J$. At a given time,the composition of the reaction mixture is $[A] = 1/2, [B] = 2$ and $[C] = 1/2$. The reaction proceeds in the:

For the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,at equilibrium,the mole fraction of $PCl_5$ is $0.4$ and the mole fraction of $Cl_2$ is $0.3$. What will be the mole fraction of $PCl_3$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module