Question

In a system AB2(s)in equilibrium give A(g)+2B(g) on doubling the quantity of AB2 (s) at equilibrium ? Why in this question partial pressure of B remain same.

Answer 1

Answer:Partial Pressures: In a mixture of gases, it is the pressure an individual gas exerts. The partial pressure is independent of other gases that may be present in a mixture. That is why it remains unchanged

Explanation:

Answer 2

Answer:

On doubling the quantity of AB₂(s) will not affect the partial pressure of B since AB₂ is in solid-state and for solid-state activity always remains 1. So, the partial pressure of B remains the same.

Explanation:

First, let's understand what is partial pressure.

In a mixture of gases, each constituent gas has certain pressure that it exerts and that certain pressure is independent of other gases that may be present in a mixture. And this certain pressure of the gas is called partial pressure of the gas.

$P_x=n_xP_t_o_t_a_l$             (1)

Px=partial pressure of the gas

nx= number of moles of the gas

Ptotal = partial pressure of all the constituent pressure present in the container

The reaction given in the question is,

$AB_2(s) \rightarrow A(g)+2B(g)$         (2)

On doubling the quantity of AB₂(s) will not affect the partial pressure of B since AB₂ is in solid-state and for solid-state activity always remains 1.

Hence, the partial pressure of B remains the same.