# Find kp

b) A(g) + 2B(g)

2C(g)

In the above reaction in a closed rigid vessel initially 1.0 mol A(g) and 2.0mol B(r) were taken and when the temperature was raised to 600K, 0.4mol C(j) appeared at equilibrium. In this case the total compressibility of the block is 2.8 × 105Pa at 600K to calculate the above equilibrium constant kp. To predict the above equilibrium constant kc at 600K.
Original post by Miss kp
b) A(g) + 2B(g)

2C(g)

In the above reaction in a closed rigid vessel initially 1.0 mol A(g) and 2.0mol B(r) were taken and when the temperature was raised to 600K, 0.4mol C(j) appeared at equilibrium. In this case the total compressibility of the block is 2.8 × 105Pa at 600K to calculate the above equilibrium constant kp. To predict the above equilibrium constant kc at 600K.

First and foremost, let’s clarify a few things:

The equation should presumably be written as

A (g) + 2B (g) <=> 2C (g)

And the compressibility should presumably be taken to read as 2.8 x 10^5 Pa.

What have you tried so far?