Question

In a reaction PCl₅⇌PCl₃+Cl₂ degree of dissociation is 30%. If initial moles of PCl₅ is one then total moles at equilibrium is: a. 1.3 b. 0.7 c. 1.6 d. 1.0

Answer 1

The total moles at equilibrium for the dissociation of PCl₅ into PCl₃ and Cl₂ with a 30% degree of dissociation, starting with 1 mole of PCl₅, is 1.3 moles. The correct answer is (a) 1.3.

Step-by-step explanation:

The question asks for the total moles at equilibrium after the dissociation of PCl₅ into PCl₃ and Cl₂, with a degree of dissociation of 30%. If the initial moles of PCl₅ is one, we can calculate the moles of PCl₃ and Cl₂ produced, as well as the remaining moles of PCl₅ at equilibrium using the degree of dissociation.

For PCl₅ dissociating into PCl₃ and Cl₂:

PCl₅ → PCl₃ + Cl₂

Initial moles: 1----0----0

Change in moles: -0.3---+0.3---+0.3

Moles at equilibrium: 0.7---0.3---0.3

Therefore, the total moles at equilibrium = moles of PCl₅ + moles of PCl₃ + moles of Cl₂ = 0.7 + 0.3 + 0.3 = 1.3. So the correct answer is (a) 1.3.

Answer 2

The correct answer is d.) The total moles at equilibrium is 1.0.

Step-by-step explanation:

The degree of dissociation of a reaction is defined as the fraction of the initial moles of a substance that dissociates into its components at equilibrium. In this reaction, PCl5 dissociates into PCl3 and Cl2.

Given that the degree of dissociation is 30%, it means that 30% of the initial moles of PCl5 dissociated into PCl3 and Cl2.

Therefore, the total moles at equilibrium would be the sum of the moles of PCl5 that did not dissociate (70% of the initial moles) and the moles of PCl3 and Cl2 that were formed through dissociation. Since the initial moles of PCl5 is one, the total moles at equilibrium would be 0.7 + 0.3 = 1.0.