Question

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


`S_2 (g) + C (s) \rightleftharpoons CS_2 (g)`;
`K_c` = 9.40 at 900 K
How many grams of CS₂ (g) can be prepared by heating 15.5 mol S₂ (g) with excess carbon in a 8.30 L reaction vessel held at 900 K until equilibrium is attained?

Answer 1

1067 g

Step-by-step explanation:

The equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants in a reversible reaction. The equilibrium constant, Kc, is calculated for a generic reaction:

aA + bB ⇄ cC + dD

`Kc = ([C]^c*[D]^d)/([A]^a*[B]*b)`

Those concentrations are a simplification of the activity of the compounds. Solids and water have activity equal to 1, so they're not placed in the equation. For the reaction given, let's do an equilibrium chart, knowing that the concentration is the number of moles divided by the volume:

[S₂]initial = 15.5/8.30 = 1.8675 M

S₂(g) + C(s) ⇄ CS₂(g)

1.8675 0 Initial

-x +x Reacts (stechiometry is 1:1)

1.8675 - x x Equilibrium

Thus

Kc = x/(1.8675-x)

9.40 = x/(1.8675-x)

x = 17.5545 - 9.40x

10.40x = 17.5545

x = 1.6879 M

Then, the number of moles of CS₂ formed is the concentration multiplied by the volume:

n = 1.6879*8.30

n = 14.01 mol

The molar mass of it is 76.14 g/mol, and the mass is the molar mass multiplied by the number of moles:

m = 76.14*14.01

m = 1067 g