Question

Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equation is K940 at 900 K How many grams of CSzg) can be prepared by heating 10.9 moles of S(g) with excess carbon in a 645 L reaction vessel held at 900 K until equilibrium is attained? Number

Answer 1

S2(g)+C(s) <---------->CS2(g)

KC= [CS2]/[S2][C]

since the activity of C is unity since it is a solid

KC= [CS2]/[S2]= 9.40

initial concentration of S2 = moles/volume = 10.9/6.45=1.69

let x= drop in concentration of S2 to reach equilibrium.

At Equilibrium, [CS2]=x and [S2]= 1.69-x

Kc= x/(1.69-x)= 9.4

x= 9.4*1.69-9.4x

10.4x= 9.4*1.69

x= 1.53M

moles of CS2 formed= 1.53Moles/L* 6.45 L=9.87 moles

mass of CS2 formed = moles* molar mass of CS2 =9.87* 76 gm =750 gm