At equilibrium, 10.2g CS2 is present. Calculate Kc.

Question

At a certain temperature, 0.322 mol CH4 and 0.736 mol H2S are placed in a 2.00L container.

CH4(g)+2H2S(g)-->CS2(g)+4H2(g)

Answer 1

No. of moles of CH4 = 0.322 moles, [CH4] = 0.322 moles /1.5 L = 0.2146M

No. of moles of H2S = 0.736 moles,,

Volume of a container: V = 1.5 L

mass of CS2 at equilibrium is 10.2 g

Initial concentration of CH4 = number of moles of CH4/volume of a container = 0.322 moles /1.5L = 0.2146M

Initial concentration of H2S = number of moles of H2S/volume of a container = 0.736 moles /1.5L = 0.490 M

Molar mass of CS2 = 76.139 g/mol

No. of moles of CS2 at equilibrium = w / mw = 10.2g / 76.139g/mol = 0.1339 mol then, [CS2] = 0.1339mol/1.5L = 0.0892M

[H2S] = 0.490M-2x = 0.49M-2x0.0892M = 0.311M, where X= = 0.0892M.

[H2]= 4x0.0892M =0.356M

The equilibrium constant for the above reaction is as follows:

Kc = [CS2][H2]⁴

[CH4][H2S]²

^{Substitute the above vales into the equation}

^{Kc = 0.0892 x (0.356)4}

0.2146x(0.311)²

Kc = 0.00143/ 0.02075 = 0.06891M

Therefore, Kc = 0.06891M