Equilibrium reaction: M_{2} + N_{2} ⇌ 2MN

**When dealing with equilibrium and K _{c}:**

• **K _{c} **→ equilibrium units are in molarity

•

$\overline{){{\mathbf{K}}}_{{\mathbf{c}}}{\mathbf{=}}\frac{\mathbf{products}}{\mathbf{reactants}}{\mathbf{=}}\frac{{\left[\mathrm{MN}\right]}^{\mathbf{2}}}{\left[{M}_{2}\right]\left[{N}_{2}\right]}}$

Solve for **K**_{c} using diagram A:

${\mathbf{K}}_{\mathbf{c}}\mathbf{=}\frac{{\left[\mathbf{0}\mathbf{.}\mathbf{04}\right]}^{\mathbf{2}}}{\left[\mathbf{0}\mathbf{.}\mathbf{02}\right]\left[\mathbf{0}\mathbf{.}\mathbf{02}\right]}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{c}}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}}{\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}}$

**K _{c} = 4.0**

**ICE Chart for diagram B:**

For the reaction M_{2} + N_{2} ⇌ 2MN, scene A represents the mixture at equilibrium, with M black and N orange. If each molecule represents 0.10 mol and the volume is 1.0 L, how many moles of each substance will be present in scene B when that mixture reaches equilibrium?

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Based on our data, we think this problem is relevant for Professor E. Lai Hing's class at OAKWOOD.

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Our data indicates that this problem or a close variation was asked in Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition. You can also practice Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition practice problems.