**Equilibrium equation**: SCl_{2}(g) + 2C_{2}H_{4}(g) ⇌ S(CH_{2}CH_{2}Cl)_{2}(g)

**When dealing with equilibrium:**

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

•

**K _{c}** is an equilibrium expression:

$\overline{){{\mathbf{K}}}_{{\mathbf{c}}}{\mathbf{=}}\frac{\mathbf{products}}{\mathbf{reactants}}{\mathbf{=}}\frac{\left[S{\left({\mathrm{CH}}_{2}{\mathrm{CH}}_{2}\mathrm{Cl}\right)}_{2}\right]}{\mathbf{\left[}{\mathbf{SCl}}_{\mathbf{2}}\mathbf{\right]}{\mathbf{\left[}{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}\mathbf{\right]}}^{\mathbf{2}}}}$

**K _{p} and K_{c} are related:**

$\overline{){{\mathbf{K}}}_{{\mathbf{p}}}{\mathbf{=}}{{\mathbf{K}}}_{{\mathbf{c}}}{\left(\mathbf{RT}\right)}^{\mathbf{\u2206}\mathbf{n}}}$

** ****ICE Chart**

At equilibrium: **[S(CH _{2}CH_{2}Cl)_{2}] = x = 0.350 M**

**[****SCl**] = 0.675 - 0.350 = 0.325 M_{2}**[****C**] = 0.973 - 2(0.350) = 0.273 M_{2}H_{4}

$\overline{){{\mathbf{K}}}_{{\mathbf{c}}}{\mathbf{=}}\frac{\left[S{\left({\mathrm{CH}}_{2}{\mathrm{CH}}_{2}\mathrm{Cl}\right)}_{2}\right]}{\mathbf{\left[}{\mathbf{SCl}}_{\mathbf{2}}\mathbf{\right]}{\mathbf{\left[}{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{4}}\mathbf{\right]}}^{\mathbf{2}}}}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{c}}\mathbf{=}\frac{\left[\mathbf{0}\mathbf{.}\mathbf{350}\right]}{\left[\mathbf{0}\mathbf{.}\mathbf{325}\right]{\left[\mathbf{0}\mathbf{.}\mathbf{273}\right]}^{\mathbf{2}}}$

**K _{c} = 14.45**

A toxicologist studying mustard gas, S(CH_{2}CH_{2}Cl)_{2}, a blistering agent, prepares a mixture of 0.675 M SCl_{2} and 0.973 M C_{2}H_{4} and allows it to react at room temperature (20.0°C):

SCl_{2}(g) + 2C_{2}H_{4}(g) ⇌ S(CH_{2}CH_{2}Cl)_{2}(g)

At equilibrium, [S(CH_{2}CH_{2}Cl)_{2}] = 0.350 M. Calculate K_{p}.

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