Consider bromine monochloride, an interhalogen compound. It is formed by the rea
ID: 484366 • Letter: C
Question
Consider bromine monochloride, an interhalogen compound. It is formed by the reaction between red-orange bromine vapor and yellow chlorine gas; BrCl is itself a gas. The reaction is endothermic. Write the chemical equation for the formation of BrCl, using simplest whole number coefficients. Write the equilibrium expression for Kc At 400 degree C, after the reaction reached equilibrium, the mixture contained 0.82 M BrCl, 0.20 M Br2, and 0.48 M Cl2. Calculate Kc for the reaction Write the expression for Kp What is Kp at 400 degree C? Initially, a 2.00 L flask contains Cl2 with a partial pressure of 0.51 atm and Br2 with a partial pressure of 0.34 atm. After equilibrium is established, the partial pressure of BrCl is 0.46 atm. Calculate the equilibrium partial pressures of Cl2 and Br2. Initially, a 2.00 L flask contains 0.15 mol of each gas are placed in the same container. In what direction will the reaction proceed?Explanation / Answer
The reaction between Bromine vapor and chlorine gas is
a)Br2(g)+Cl2(g)--à2BrCl ( the simplest whole numbers are 1 each for Br2, Cl2 and 2 for BrCl)
b) Kc= Equilibrium constant = [BrCl]2/ [Br2][Cl2]
c) with the given values Kc= (0.82)2/( 0.2*0.48) = 7
d) Kp= Kc*(RT) deltan, deltan= change in no of moles = moles of products-moles of reactants
deltan= 0
e) hence at 400 deg.c, Kp=KC= 7
f) partial pressure of Cl2= 0.51 atm and Br2= 0.34 atm
let x= drop in partial pressure of Br2 to reach equilibrium
at equilibrium [PCl2] = 0.51-x, [PBr2] =0.34-x, [PBrCl] =2x
given 2x= 0.46, x=0.23, at equilibrium PCl2= 0.51-0.23= 0.28 atm, PBr2= 0.34-0.28=0.06 atm and PBrCl=2*0.06=0.12 atm
g) The reaction is endothermic, as per the reaction Br2+Cl2-à2BrCl
initially [Br2]= 0.15/2 =0.075 M =[Cl2], [BrCl]=0.15/2=0.075
let x= drop in concentration to reach equilibrium
at Equilibrium [Br2] = [Cl2] = 0.075-x, [BrCl] =0.075+2x
Kc= (0.075+2x)2/ (0.075-x)2= 7, taking square root (0.075+2x)= 2.64*(0.075-x)
0.075+2x= 2.64*0.075-2.64x
Hence x*(2+2.64)= 0.075*(2.64-1)= 0.075*1.64, x= 0.075*1.64/4.64 = 0.0265 moles
At equilibrium [Br2] =[Cl2] = 0.075-0.0265=0.0485M, [BrCl]= 0.075+0.0265= 0.1015. So the reaction proceeds to product side.