A(s)=2B(g) C(g) The above equilibrium was established by taking A(s) only. At equilibrium B is removed so that it's partial pressure at new equilibrium becomes 1/3 of original total pressure. Ratio of total pressure at new equilibrium and at initial equilibrium will be?

Question

Answer

Ans is 1/3 bcz after removing b only c left for eqlbm cons.so it's partial pressure is equal to total pressure at that time

Answer

The given equilibrium reaction is:
A(s) ⇌ 2B(g) + C(g)

The equilibrium was established by taking only A(s). This means that initially, there was only A(s) present and no B(g) or C(g).

Now, B(g) is removed from the system, and we need to determine the new equilibrium conditions.

Let's analyze the changes that occur when B(g) is removed:

1. Removal of B(g):
- Since B(g) is being removed, the reaction will shift to the left to compensate for the loss of B(g).
- As a result, the concentration of A(s) will increase, and the concentrations of B(g) and C(g) will decrease.

2. Partial pressure of B at new equilibrium:
- We are given that the partial pressure of B at the new equilibrium is 1/3 of the original total pressure.
- This means that the concentration of B(g) at the new equilibrium is 1/3 of the original concentration.

Now, let's determine the ratio of total pressure at the new equilibrium to the initial equilibrium:

1. Initial equilibrium:
- At the initial equilibrium, only A(s) is present.
- Since A(s) is a solid, its concentration does not change, and its partial pressure is negligible.
- Therefore, the total pressure at the initial equilibrium is solely due to B(g) and C(g).

2. New equilibrium:
- At the new equilibrium, B(g) and C(g) are present.
- We know that the partial pressure of B at the new equilibrium is 1/3 of the original total pressure.
- Since the initial equilibrium only consisted of B(g) and C(g), the total pressure at the new equilibrium will be 1/3 of the initial total pressure.

In conclusion, the ratio of the total pressure at the new equilibrium to the initial equilibrium is 1/3.

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