2 NOCl_{(g)} rightleftharpoons 2 NO_{(g)} + C | vedclass

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(B) The reaction is: $2 NOCl_{(g)} \rightleftharpoons 2 NO_{(g)} + Cl_{2(g)}$Initial concentration: $[NOCl] = 2.0 \ M$,$[NO] = 0 \ M$,$[Cl_2] = 0 \ M$

Vedclass · Accepted Answer

$125$

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(B) The reaction is: $2 NOCl_{(g)} ightleftharpoons 2 NO_{(g)} + Cl_{2(g)}$Initial concentration: $[NOCl] = 2.0 \ M$,$[NO] = 0 \ M$,$[Cl_2] = 0 \ M$At equilibrium: $[NOCl] = (2 - x) \ M$,$[NO] = x \ M$,$[Cl_2] = \frac{x}{2} \ M$Given that at equilibrium,$[NO] = 0.4 \ M$,so $x = 0.4 \ M$.Therefore,$[NOCl]_{eq} = 2 - 0.4 = 1.6 \ M$ and $[Cl_2]_{eq} = \frac{0.4}{2} = 0.2 \ M$.The equilibrium constant $K_c$ is given by:$K_c = \frac{[NO]^2 [Cl_2]}{[NOCl]^2} = \frac{(0.4)^2 	imes (0.2)}{(1.6)^2}$$K_c = \frac{0.16 	imes 0.2}{2.56} = \frac{0.032}{2.56} = 0.0125$$K_c = 125 	imes 10^{-4}$

$2 NOCl_{(g)} \rightleftharpoons 2 NO_{(g)} + Cl_{2(g)}$
In an experiment,$2.0 \ mol$ of $NOCl$ was placed in a $1 \ L$ flask and the concentration of $NO$ after equilibrium was established,was found to be $0.4 \ mol/L$. The equilibrium constant at $30^{\circ} C$ is $....... \times 10^{-4}$.

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$2 NOCl_{(g)} \rightleftharpoons 2 NO_{(g)} + Cl_{2(g)}$In an experiment,$2.0 \ mol$ of $NOCl$ was placed in a $1 \ L$ flask and the concentration of $NO$ after equilibrium was established,was found to be $0.4 \ mol/L$. The equilibrium constant at $30^{\circ} C$ is $....... \times 10^{-4}$.