I have 2 questions below. Please help with step-by-step even if you can only do

Question

I have 2 questions below. Please help with step-by-step even if you can only do one.

#1:

Hydrogen sulfide decomposes according to the following equation, for which Kc = 0.00064 at a given temperature.
2 H2S(g) <=> 2 H2(g) + S2(g)
If 7.02mol of H2S is placed in a 3.0 L container, what is the equilibrium concentration of H2(g)?

ANSWER: 0.19138

#2:

Ammonium hydrogen sulfide decomposes according to the following reaction, for which Kp = 0.114 at a given temperature. If 55.0 g of NH4HS is sealed inside a 5.00 L container at this temperature, calculate the partial pressure of NH3(g) at equilibrium:
NH4HS(s) <=> NH3(g) + H2S(g)

ANSWER: 0.3376

Explanation / Answer

1)
2 H2S(g) <=> 2 H2(g) + S2(g)

-2x--------------- +2x --------+x

[H2S] = 7.08mol/3L = 2.36M

Kc = 6.4*10^-3 = ([H2]^2*[S2])/(H2S)^2

Kc = (2x)^2 * x / (2.36 - 2X)^2 = 0.00064

x = .091

H2 = 2*x = .182 M.

2)
First, let's write the equilibrium partial pressure using "x" as the change in partial pressure. NH4HS doesn't have a partial pressure because it is a solid, so it will be a constant in the equation, or a value of 1.


NH4HS(s) = (constant)
NH3 = x
H2S = x

Kp = P[NH3]*P[H2S]/1 = 0.114

substitute equilibrium values from the table

0.114= x^2

x = 0.337 atm = partial pressure of NH3

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