Question

At a certain temperature, 0.332 mol of CH4 and 0.953 mol of H2S is placed in a 2.00 L container.

At equilibrium, 14.3 g of CS2 is present. Calculate Kc.​

Answer 1

Given,

Number of moles of CH₄ = 0.332 mol

Number of moles of H₂S = 0.953 mol

Volume of container = 2.00 L

At equilibrium, Mass of CS₂ = 14.3 g

Now, the balanced reaction between CH₄ and H₂S is,

CH₄(g) + 2H₂S(g) 4H₂(g) + CS₂(g)

Now,

Calculating the concentration of CH₄ and H₂S,

[CH₄] = 0.332 mol / 2.00 L = 0.166 M

[H₂S] = 0.953 mol / 2.00 L = 0.4765 M

Calculating the number of moles of CS₂ and then the concentration of CS₂ at equillibrium,

= 14.3 g CS₂ x ( 1 mol / 76.139 g)

= 0.1878 mol / 2.00 L

= 0.09390 M

Now, Drawing an ICE chart for the balanced reaction,

	CH4	2H2S	4H2	CS2
I(M)	0.166	0.4765	0	0
C(M)	-x	-2x	+4x	+x
E(M)	0.166-x	0.4765-2x	4x	x

Now, We know,

x = [CS₂] = 0.09390 M

[CH₄] = 0.166 -x = 0.166-0.09390 = 0.07209 M

[H₂S] = 0.4765 - 2x = 0.4765 - 2 x 0.09390 = 0.2886 M

[H₂] = 4x = 4 x 0.09390 = 0.3756 M

Now, the K_c expression for the given reaction is,

K_c = [H₂]⁴ [CS₂] / [CH₄] [H₂S]²

K_c = [0.3756]⁴ [0.09390] / [0.07209] [0.2886]²

K_c = 0.311