Question

The equilibrium expression for a reaction is `"K"_("eq") = ("[H"^+"]"^6)/("[Bi"^(2+)"]"^2["H"_2"S"]^3)` Which of the following could be the reaction?

A.
6H+(aq) + BiS(s) 2Bi2+(aq) + 3H2S(g)

B.
2Bi2+(aq) + 3H2S(aq) Bi2S3(s) + 6H+(aq)

C.
6H+(aq) + Bi2S3(s) 2Bi2+(aq) + 3H2S(g)

D.
2Bi2+(aq) + 3H2S(aq) Bi2S3(aq) + 6H+(aq)

Answer 1

Option B is correct

K = Kp /Kr

The given equation indicating, the product containing 6 moles of proton whereas the reactant contains 2 mole of bismuth and 3 mole of hydrogen sulphide.

Hence, in reaction B there are 2 mole of bismuth and 3 mole of hydrogen sulphide reacting to produce 6 moles of proton. whereas the concentration of Bi2S3 is not considered as it is present in solid phase.

Answer 2

Answer:

2Bi2+(aq) + 3H2S(aq)↔ Bi2S3(s) + 6H+(aq)

Explanation:

In general for a hypothetical reaction:

x(Reactants) ↔ y(Products)

the equilibrium constant, Keq is given by the ratio of the products to that of the reactants raised to the appropriate coefficients.

$Keq = \frac{[Products]^{y} }{[Reactants]^{x} }$

The given Keq is:

$Keq = \frac{[H+]^{6}}{[Bi2+]^{2} [H2S]^{3} }$

This implies that the reactants are: H2S (aq) and Bi2+(aq)

Products are: H+(aq)

Therefore the reaction with the appropriate coefficients would be:

2Bi2+(aq) + 3H2S(aq)↔ Bi2S3(s) + 6H+(aq)

Since the activity of solids = 1, Bi2S3 is not included in Keq