Problem: 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.Give your answer to 2 decimal places.

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

SCl_2(g) + 2 C₂H_4(g) ↔ S(CH₂CH₂Cl)_2(g)

When dealing with equilibrium and K_c:

• K_c → equilibrium units are in molarity 
• K_c is an equilibrium expression:

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

▪ 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:    SCl_2(g) + 2 C₂H_4(g) ↔ S(CH₂CH₂Cl)_2(g)