# Problem: A toxicologist studying mustard gas, S(CH2CH2Cl)2, a blistering agent, prepares a mixture of 0.675 M SCl2 and 0.973 M C2H4 and allows it to react at room temperature (20.0°C):SCl2(g) + 2C2H4(g) ⇌ S(CH2CH2Cl)2(g)At equilibrium, [S(CH2CH2Cl)2] = 0.350 M. Calculate Kp.

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###### FREE Expert Solution

Equilibrium equation: SCl2(g) + 2C2H4(g) ⇌ S(CH2CH2Cl)2(g)

When dealing with equilibrium:

Kc → equilibrium units are in molarity
Kp → equilibrium units in terms of pressure

Kc is an equilibrium expression:

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

Kp and Kc are related:

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

ICE Chart At equilibrium: [S(CH2CH2Cl)2] = x = 0.350 M

• [SCl2] = 0.675 - 0.350 = 0.325 M
• [C2H4] = 0.973 - 2(0.350) = 0.273 M

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

Kc = 14.45 ###### Problem Details

A toxicologist studying mustard gas, S(CH2CH2Cl)2, a blistering agent, prepares a mixture of 0.675 M SCl2 and 0.973 M C2H4 and allows it to react at room temperature (20.0°C):
SCl2(g) + 2C2H4(g) ⇌ S(CH2CH2Cl)2(g)
At equilibrium, [S(CH2CH2Cl)2] = 0.350 M. Calculate Kp.