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.

A toxicologist is a scientist who is well-versed in a variety of scientific fields, including biology and chemistry. Typically, they deal with chemicals and other compounds to ascertain whether they are toxic or damaging to people, other living things, or the environment.

We must determine the K p of the equilibrium reaction illustrated below in this reaction.

First, note the values that are provided.

2 (g)+2C 2 H 4 (g)⇌S(CH 2 CH 2Cl) 2(g)

T=20∘ C=293.15 K

[{SCl}_2]_i=0.675 \ \{M}

[SCl 2] i =0.675 M

{C}_2{H}_4]_i=0.973{M}

[C 2H 4] i=0.973 M

[{S(CH}_2{CH}_2{Cl)}_2]_{eq}=0.350 {M}

[S(CH 2CH 2Cl) 2] eq =0.350 M

We can find xx since we are aware of the concentration of S(CH 2CH 2Cl) 2S(CH 2 CH 2 Cl) at equilibrium.

[S(CH} 2{CH} 2{Cl)} 2]

_{eq}=x=0.350

[S(CH 2 CH 2 Cl)]

eq \s =x=0.350

This means that x=0.350x=0.350. Determine the reactant concentration at equilibrium by solving.

[ SCl 2]

[SCl 2] _eq &=0.675 -x &=0.675-0.350_eq&=0.325 M [SCl 2 ]eq

[SCl 2]=0.675−x =0.675−0.350 \s=0.325 M

[C2 H4 ] eq[C 2 H 4 ] eq=0.973−2x=0.973−2(0.350)=0.273 M

Lastly solve for the K_pK of the reaction.

KpK p=K c (RT) Δn=(14.45)(0.0821×293.15) (1−3)= 0.0249

Kp=0.0249

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