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 12.6 moles of S2(g) with excess carbon in a 6.95 L reaction vessel held at 900 K until equilibrium is attained?

Answer 1

Reaction between Sulphur and charcoal to give Carbon disulphide

S₂ (g) + C (s) <===> CS₂ (g)

Equilibrium constant Kc=[CS₂]/([S₂][C])=9.40 at 900 K (given in question)

Given number of moles of Sulphur= 12.6

Volume of vessel=6.95 L

Concentration of Sulphur=number of moles/Volume in L=12.6 moles/6.95 L1.81 M

From the balanced chemical equation above we can say that

1 mole of Sulphur will react with one mole of Carbon

So 12.6 moles of S will react with 12,6 moles of C (given there is excess C present)

So 1.81 M Sulphur reacts with 1.81 M Carbon

S₂ (g) + C (s) <===> CS₂ (g)

Initial concentration(M) 1.81 M 1.81 0   

Change(M) -x -x +x

Equilibrium concentration(M) 1.81-x 1.81-x x

Kc=[CS₂]/([S₂][C])=9.40

9.40=x/(1.81-x)(1.81-x)

9.40(1.81-x)²=x

9.40(3.2761+x²-3.62x)=x

9.4x²-34.028x+30.7953=x

9.4x²-34.028x-x+30.7953=0

9.4x²-35.028x+30.7953=0

so taking positive value

x={-(-35.028){(-35.028)²-4(9.4)(30.7953)}^1/2}/2(9.4)

x={+35.028(69.0975)^1/2}/18.8

x=(+35.0288.31)/18.8

x=2.305 M or 1.42 M

So either 2.305 M of CS₂ or 1.42 M of CS₂ can be prepared

Molar mass of CS₂=Molar mass of C+2xMolar mass of S

=12+2x32=12+64 g/mol=76 g/mol

number of moles=Molarityx Volume in L

so either 2.305x6.95 moles or 1.42x6.95 moles of CS₂ will be produced

either 16.01975 moles or 9.869 moles of CS₂ will be produced

Now 12.6 moles of S cannot produce 16.01975 moles of CS₂ according to the balanced chemical equation

So 9.869 moles of CS₂ will be produced

Mass of CS₂ produced=number of molesxmolar mass=9.869 molesx76 g/mol=750.044 g of CS2 will be produced