Question

Nitrosyl Chloride, NOCl(g) decomposes to form nitrogen monoxide gas and chloride gas according to the following equation.
2NOCl(g) double arrow 2NO(g) + Cl2(g)
At a certain temperature, the equilibrium constant is 1.60 x 10-6 (that's supposed to be an exponent). Calculate the equilibrium concentrations of all gases if 0.800 moles of NOCL(g) are placed in a 2.00 L container.

Answer 1

Question #57686, Chemistry, Other

Nitrosyl Chloride, $\mathrm{NOCl}_{(g)}$ decomposes to form nitrogen monoxide gas and chloride gas according to the following equation.

$$2 \mathrm{NOCl}_{(g)} \leftrightarrow 2 \mathrm{NO}_{(g)} + \mathrm{Cl}_{2(g)}$$

At a certain temperature, the equilibrium constant is $1.60 \times 10^{-6}$ (that's supposed to be an exponent). Calculate the equilibrium concentrations of all gases if 0.800 moles of $\mathrm{NOCl}_{(g)}$ are placed in a $2.00\,\mathrm{L}$ container.

Answer:

$$K_{eq} = \frac{[\mathrm{NO}]^2 [\mathrm{Cl}_2]}{[\mathrm{NOCl}]^2}$$

0.800 mol NOCl / 2.0 L = 0.400 M NOCl

$$\begin{array}{l} 1.60 \times 10^{-6} = [\mathrm{NO}]^2 [\mathrm{Cl}_2] / [\mathrm{NOCl}]^2 = (2x)^2(x) / (0.4 - 2x)^2 = 4x^3 / 0.4^2 \\ x = 4 \times 10^{-3} \end{array}$$

From here:

$$[\mathrm{NOCl}] = 0.4 - 2 \cdot (4 \times 10^{-3}) = 0.392$$ $$[\mathrm{NO}] = 2 \cdot (4 \times 10^{-3}) = 8 \times 10^{-3}$$ $$[\mathrm{Cl}_2] = 4 \times 10^{-3}$$

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