# Problem: Mustard gas, S(CH2CH2Cl)2 can be made at room temperature as shown:SCl2(g) + 2 C2H4(g) &lt;=&gt; S(CH2CH2Cl)2(g)In a particular experiment, 0.816 M of SCl2 and 0.288 M C2H4 were mixed. At equilibrium,{S(CH2CH2Cl)2(g)} was 0.187 M.Calculate the value of Kc for the reaction.Give your answer to 2 decimal places.

###### FREE Expert Solution

We are being asked to calculate the equilibrium constant, Kc for the given equilibrium reaction:

SCl2(g) + 2 C2H4(g) ↔ S(CH2CH2Cl)2(g)

When dealing with equilibrium and Kc:

Kc → equilibrium units are in molarity
Kc is an equilibrium expression:

$\overline{){{\mathbf{K}}}_{{\mathbf{c}}}{\mathbf{=}}\frac{\mathbf{products}}{\mathbf{reactants}}}$

only aqueous and gaseous species are included in the equilibrium expression
▪ the coefficient of each compound in the reaction equation will be the exponent of the concentrations in the equilibrium expression

Given:    SCl2(g) + 2 C2H4(g) ↔ S(CH2CH2Cl)2(g)

###### Problem Details

Mustard gas, S(CH2CH2Cl)2 can be made at room temperature as shown:

SCl2(g) + 2 C2H4(g) <=> S(CH2CH2Cl)2(g)

In a particular experiment, 0.816 M of SCl2 and 0.288 M C2H4 were mixed. At equilibrium,

{S(CH2CH2Cl)2(g)} was 0.187 M.

Calculate the value of Kc for the reaction.