Question

Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equation is

S2(g)+C(s)=CS2(g)             Kc=9.40 at 900k

How many grams of CS2(g) can be prepared by heating 18.0 moles of S2(g) with excess carbon in a 8.05 L reaction vessel held at 900 K until equilibrium is attained?

Answer 1

consider the given equation

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

we know that

solids are not considered for equilibrium constant

so

Kc = [CS2] / [S2]

now

we know that

concentration = moles / volume (L)

so

initially

[S2] = 18 / 8.05 = 2.236 M

now

S2 + C (s) ---> CS2

using ICE table

initial conc of S2 , CS2 are 2.236 , 0

change in conc of S2 , CS2 are -x , x

equilibrium conc of S2 , CS2 are 2.236 -x , x

now

Kc = [CS2] / [S2]

9.4 = [ x ] / [2.236 -x ]

x = 2.021

so

at equilibrium

[CS2] = 2.021

we know that

moles = molarity x volume (L)

so

moles of CS2 = 2.021 x 8.05

moles of CS2 = 16.26905

now

mass = moles x molar mass

so

mass of CS2 = 16.26905 x 76

mass of CS2 = 1236.4

so

1236.4 grams of CS2 can be prepared