87.8k views
5 votes
2HI(g)→H₂(g)+I₂(g)

The equilibrium constant of the above reaction is 6.4 at 300 K. If 0.25 mole each of H₂ and I₂ are added to the system, the equilibrium constant will be
A 0.8
B 3.2
C 1.6
D 6.4

User ZealousWeb
by
7.9k points

1 Answer

1 vote

Final answer:

The question seeks to calculate the equilibrium partial pressures of gases in the decomposition of HI to H₂ and I₂. By applying the equilibrium constant expression to the given starting pressure and Kc at 698K, the partial pressures at equilibrium can be found.

Step-by-step explanation:

The question deals with a chemical equilibrium scenario involving the reaction where hydrogen iodide (HI) decomposes into hydrogen gas (H₂) and iodine gas (I₂). The reaction, represented as 2HI(g) → H₂(g) + I₂(g), has an equilibrium constant Kc at a specific temperature. To determine the partial pressures of the gases at equilibrium, we apply the equilibrium constant expression Kc = (PH₂PI₂) / PHI2, where P denotes the partial pressures of the gases. In this scenario, given a starting partial pressure of HI and the value of Kc, one would calculate the change in concentration of HI to find the concentrations of all substances at equilibrium.

At a temperature of 698K, when the reaction starts with HI at a partial pressure of 1 atm and reaches equilibrium, we can apply the equilibrium constant and stoichiometry to calculate the partial pressures for H₂, I₂, and HI at equilibrium. This involves setting up an ICE (Initial, Change, Equilibrium) table and solving for the unknown pressures.

User Pranay Dutta
by
8.6k points