Consider the following reaction at equilibrium: 2H2( g )+S2( g )2H2S( g )+heat I

Question

Consider the following reaction at equilibrium:

2H2(g)+S2(g)2H2S(g)+heat

In a 10.0-L container, an equilibrium mixture contains 2.20 g of H2, 10.9 g of S2 and 68.1 g of H2S.

Part A What is the numerical Kc value for this equilibrium mixture?

Part B If more H2 is added to the equlibrium mixture, how will the equilibrium shift?

Part C How will the equilibrium shift if the mixture is placed in a 5.00-L container with no change in temperature?

Part D If a 6.00-L container has an equilibrium mixture of 0.250 mol of H2 and 2.40 mol of H2S, what is the [S2] if temperature remains constant?

Please please help!!

Explanation / Answer

2H2(g)+S2(g) <---> 2H2S(g)+heat

at equilibrium,

concentration of H2 = (2.2/2)*(1/10) = 0.11 M

concentration of S2 = (10.9/64)*(1/10) = 0.017 M

concentration of H2S = (68.1/34)*(1/10) = 0.2 M

part A

Kc = [H2S]^2/[H2]^2[S2]

    = (0.2^2/(0.11^2*0.017))

    = 194.5

part B

equilibrium shift to right, towards product side.

part C

volume of container is decreased. results partal pressures of components increases,so that equilibrium shift towards less no of mols side(products side).

equilibrium shift to right.

part D

concentration of H2 = 0.25/6 =

concentration of H2s = 0.24/6 =

concentration of [S2] = ?

Kc = [H2S]^2/[H2]^2[S2]

194.5 = ((0.24/6)^2/((0.25/6)^2*X))

X = [s2] = 0.00474 M

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