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  • <img alt="2 \mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g})" src="https://entrancecorner.oncodecogs.com/gif.latex?2%20%5Cmathrm%7BNOCl%7D%28%5Cma

2 \mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g})
In an experiment, 2.0 moles of \mathrm{NOCl} was placed in a one-litre flask and the concentration of \mathrm{NO} after equilibrium established, was found to be 0.4 \mathrm{~mol} / \mathrm{L}. The equilibrium constant at 30^{\circ} \mathrm{C} is____________\times 10^{-4}.

Option: 1

125


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

The equlibrium can be solved as 

              \mathrm{2 NOCl ( g ) \quad \rightleftharpoons \quad 2 NO ( g )\quad+\quad Cl _{2}( g )}

\mathrm{t=0 \quad\quad 2 \text{M} \quad\quad \quad\quad \quad \quad 0 \quad\quad\quad \quad \quad \quad 0}

\mathrm{t=t_{eq} \quad (2-x) \text{M} \quad \quad \quad \quad xM \quad \quad \quad \quad \frac{x}{2}M} 

\mathrm{t=t_{eq} \quad 1.6 \text{M} \quad \quad \quad \quad 0.4M \quad \quad \quad \quad 0.2M}  \left (\because \mathrm{x=0.4 \mathrm{M} \text { at } \mathrm{eq}^{\mathrm{m}}} \right )

\mathrm{K_{c}=K_{e q}=\frac{(0.4)^{2} \times 0.2}{(1.6)^{2}}=\frac{0.4 \times 0.4 \times 0.2}{1.6 \times 1.6}}

\begin{aligned} &=\frac{2}{16} \times 10^{-1}=\frac{1}{8} \times 10^{-1} \\ &=0.125 \times 10^{-1} \\ &=125 \times 10^{-4} \end{aligned}

Hence, the answer is 125

Posted by

manish painkra

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