T= #298 K#,
BrCl(g) = #1.40 mol#
Volume =#12.0 L #
ΔG0[Br2(g)] = #3.11 #kJ/mol,
ΔG0[BrCl(g)] =#−0.98 #kJ/mol.]
Pls correct the equation it must be as below
#BrCl#(g)⇌#1/2Br_2#(g) + #1/2Cl_2#(g)
Or
#2BrCl#(g)⇌#Br_2#(g) + #Cl_2#(g)
Part A
Calculate ∆ G
∆ G =# 1/2# x ∆ G #(Br_2)# - ∆ G #(BrCl#)
∆ G = #½# x #(3110)- (-980)#
∆ G = #2535 J#
∆ G = - RT lnKp
lnKp = -∆ G/RT
(R = #8.314# J/mol and T = #298K#)
lnKp =# (-2535)/(8.134 x 298)# = #-1.023#
Kp = #0.3595#
Given BrCl(g)= #1.25# moles in beginning
2BrCl(g) ------> Br2(g) + Cl_2(g)
I #1.25#............................ #0# ............................... # 0 #
C # -2x #......................... # +x # ........................... #+x#
E #(1.25-2x)# ........................ # x # ...................... # x#
Kp = #[Br_2(g)] [Cl_2(g)] #/#[BrCl(g)]2#
Kp = #x2# / #(1.25 - 2x)2#
K p = #x#/#(1.25 – 2x)#
K p = #0.3595#
#(0.3595)# x# (1.25 – 2x)# = #x#
#x# = #0.4493# – #0.719x#
#x# + #0.719x# =# 0.4493#
#1.719x# = #0.4493#
#x# = #0.5033/1.719#
#x# = #0.261#
BrCl = #1.25 -2x#
BrCl = #1.25# – #(2 x 0.261)#
BrCl = #1.25 – 0.522#
BrCl = #0.727# moles
Moles of BrCl present at equillibrium = #0.727# moles
Moles of Br2 at equllibrium = #x# = #0.261# moles
Moles of Cl2 at equllibrium = #x# = #0.261# moles
