at a certain temperature, 0.351 mol CH4, a d 0.690 mol H2S are placed in 3.50 L container. CH4+2H2S<->CS2+4H2 at equilibrium, 13.9g CS2 is present. Calculate Kc.

Answer 1

CH₄ + 2H₂S <==> CS₂ + 4H₂ .. balanced equation

Kc = [CS₂][H₂]⁴ / [CH₄][H₂S]²

Best to probably use an ICE table....

CH₄ + 2H₂S <==> CS₂ + 4H₂

0.351....0.690............0.........0.........Initial moles

-x........-2x................+x.......+4x.......Change in number of moles

0.351-x..0.690-2x......x.........4x.......Equilibrium moles

We are told that there are 13,.9 g of CS₂ at equilibrium, so we can relate that to x moles as follows:

molar mass CS₂ = 76.13 g / mole

13.9 g CS₂ x 1 mol / 76.13 g = 0.183 moles CS₂ @ equilibrium = x in the ICE table.

Now we can find moles of each compound @ equilibrium by using this value and the ICE table...

moles CH₄ = 0.351 - x = 0.351 - 0.183 = 0.168 mols CH₄

moles H₂S = 0.690 - 2x = 0.690 - 0.366 = 0.324 mols H₂S

moles CS₂ = x = 0.183 mols CS₂

moles H₂ = 4x = 4 x 0.183 =0.732 mols H₂

Finally, we will divide all these values by 3.50 L to obtain CONCENTRATION (mols/L) and plug these values into the Kc equation, Kc = [CS₂][H₂]⁴ / [CH₄][H₂S]²

[CH₄] = 0.168 mol / 3.50 L = 0.0480 M

[H₂S] = 0.324 mol / 3.50 L = 0.0926 M

[CS₂] = 0.183 mol / 3.50 L = 0.0523 M

[H_2] = 0.732 mol / 3.50 L = 0.209 M

Kc = [0.0523][0.209]⁴ / [0.0480][0.0926]² = 9.98x10^-5 / 4.12x10^-4

Kc = 0.242

(be sure to check all of the math)