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