Question

Consider the reaction.

Answer 1

Answer:

0.00227 atm is the pressure of B(g)at equilibrium.

Explanation:

$2A(g)\rightleftharpoons B(g)$

The value of equilibrium constant = $K_p=7.80\times 10^{-5}$

Partial pressure of A before heating = 5.40 atm

$2A(g)\rightleftharpoons B(g)$

Initially

5.40 atm         0

at equilibrium

(5.40-2p)atm       p

The expression of an equilibrium constant can be given as

$K_p=\frac{p}{(5.40-2p)^2}$

$7.80\times 10^{-5}=\frac{p}{(5.40-2p)^2}$

Solving for p ;

p = 0.00227 atm

Partial pressure of B at 500 K = 0.00227 atm

0.00227 atm is the pressure of B(g)at equilibrium.