Question

In the gas phase equilibrium A=С+В, it was observed that the degree of dissociation of A decreased by 0.866 ratio when increasing the pressure from 1 to 2 bar. What is the value of the reaction equilibrium constant?

Answer 1

- initial mol A 1 (+8 0 0 degree of dissociation to a a Total mol = 1-x+&+Q= ita at equlibrium :: Partial pressure of different component PA = (1 - ACP Po = a and be a where pe total pressure of system Now, Kp = Pxe - ( A ) P x ( 1 ) P. Nowg kp' PA 42 1- 2 P NM ow, & «I so, we care neglect this.

Now, Now, w.ad 1 La decreases if p increase Ve Now, if we bar then x₂ = 0.866 V t if we take x = 1 (total dissociation) at Paul in x2 = 0.866 at P = 2 bar a we know, kp = x2p x²= ke ..0;}=0,= Kp (-). 1- (0.866)2 = kp (1-5) 75 Kp = pt -0.7499 = 0.2500 7 kp = 0.500)