Respuesta :
Answer:
0.00227 atm is the pressure of B(g)at equilibrium.
Explanation:
[tex]2A(g)\rightleftharpoons B(g)[/tex]
The value of equilibrium constant = [tex]K_p=7.80\times 10^{-5}[/tex]
Partial pressure of A before heating = 5.40 atm
[tex]2A(g)\rightleftharpoons B(g)[/tex]
Initially
5.40 atm 0
at equilibrium
(5.40-2p)atm p
The expression of an equilibrium constant can be given as
[tex]K_p=\frac{p}{(5.40-2p)^2}[/tex]
[tex]7.80\times 10^{-5}=\frac{p}{(5.40-2p)^2}[/tex]
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.
The pressure of B at equilibrium has been 0.00227 atm.
The equilibrium has been the position of the reaction when the concentration of reactants and products in the solution has been equal.
The balanced chemical equation for the reaction has been:
[tex]\rm 2\;A\;\rightarrow\;B[/tex]
Computation for Pressure at equilibrium:
The initial pressure of A has been given as 5.4 atm.
The initial pressure of B has been 0 atm.
The pressure of A and B at equilibrium has been given below in the ICE table attached.
The equilibrium constant, kp for the reaction, has been given as:
[tex]kp=\rm \dfrac{B}{[A]^2}[/tex]
Substituting the values for the pressure:
[tex]\rm 7.80\times\;10^-^5=\dfrac{P}{(5.4-P)^2}\\ 7.80\times\;10^-^5=\dfrac{P}{29.16-P^2}\\P=0.00227\;atm[/tex]
The pressure of B at equilibrium has been equivalent to the P, and has been 0.00227 atm.
For more information about equilibrium constant, refer to the link:
https://brainly.com/question/17960050
