Solid NH4HS is placed in a flask that has a volume of 2.80 L and contains 0.100

Question

Solid NH4HS is placed in a flask that has a volume of 2.80 L and contains 0.100 mol of NH3(g) at 25.0 degree C. The NH4HS decomposes according to the following reaction: and the equilibrium constant for this reaction is K = 0.108 at 25.0 degree C. You may assume that the NH4HS(s) is in excess and is not a limiting reagent. All gases be v What is the total gas pressure in the flask after equilibrium is established After the equilibrium pf part (a) is established, suppose that you inject an additional 0.100 mole of NH3(g). Determine the reaction quotient Q. Will this injection shift the reaction towards the reactants or products side of the equation as written above? Briefly justify with one or two sentences. What is the final equilibrium pressure of H2S(g) in the flask after the injection of NH3(g) in part (b)?

Explanation / Answer

K = [NH3] * [H2S]

0.108=[NH3] * [H2S]

1mole NH4HS gives 1mole NH3 and 1mole H2S

Since container allready contains 0.1mole NH3

At equilibrium n mole NH4HS produce (n-0.1)mole NH3 and n-0.1mole H2S

but container will be occupied by n mole NH3 and n-0.1mole H2S

0.108=n*(n-0.1)

n=0.3824 mole NH3

and 0.2824 mole H2S

total number of mole = 0.6648 mole products in gas phase in 2.8L

P = nRT/V = 0.6648*8.314*298/2.8 = 588.24N

If we inject again 0.1 mole NH3 then

Q = [NH3] * [H2S]

= (0.3824+0.1) * (0.1824) = 0.0879

Reaction quatient is 0.0879

If reaction quatient is Lesser than equilibrium constant then equilibrium shift towards reactant side

here also equilibrium shift towards reactant side

equilibrium pressure of H2S

P = nRT/V = 0.1824*8.314*298/2.8 = 161.39N partial pressure of H2S

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