Part A Cyclohexane, C 6 H 12 , undergoes a molecular rearrangement in the presen
ID: 946788 • Letter: P
Question
Part A
Cyclohexane, C6H12, undergoes a molecular rearrangement in the presence of AlCl3 to form methylcyclopentane, CH3C5H9, according to the equation:
C6H12 CH3C5H9
If K c = 0.143 at 25°C for this reaction, find the equilibrium concentrations of C6H12 and CH3C5H9 if the initial concentrations are 0.200 M and 0.150 M, respectively.
Part B
Consider the following reaction:
CuS(s) + O2(g) Cu(s) + SO2(g)
A reaction mixture initially contains 2.9 M O2. Determine the equilibrium concentration of O2 if Kc for the reaction at this temperature is 1.5.
Part C
At a certain temperature the equilibrium constant, K c, equals 0.11 for the reaction:
2 ICl(g) I2(g) + Cl2(g).
What is the equilibrium concentration of ICl if 0.25 mol of I2 and 0.25 mol of Cl2 are initially mixed in a 2.0-L flask?
Explanation / Answer
Initial [C6H12] =0.2 [ CH3C5H9] =0.15
If x= Drop in concentration of [C6H12] to reach equilibrium
At equilibrium [C6H12] =0.2-x [CH3C5H9]= 0.15+x
Kc= [CH3C5H9]/ [C6H12] =(0.2-x)/(0.15+x)= 0.143
0.2-x =0.143*(0.15+x)
0.2-x =0.143*0.15+0.143x
1.143x = 0.2-0.143*0.15=0.17855, x = 0.1562
[C6H12] =0.2-0.1562 =0.0438 [CH3C5H9]= 0.15+0.1562=0.3062
b)
Kc = [SO2]/[O2]
Initial [SO2] =0 [O2] =2.9, let x = concentration drop of O2 to reach equilibrium
At Equilibrium [SO2] = x [O2] =2.9-x
x/(2.9-x)= 1.5 , x = 1.5*(2.9-x) , x= 1.5*2.9-1.5x, 2.5x= 1.5*2.9 x = 1.5*2.9/2.5 =1.74M
At equilibrium [SO2] =1.74M and [O2] =2.9-1.74= 1.16 M
c)
Initial ]ICL ]=0 [I2]= 0.25/2 =0.125 M [Cl2] = 0.25/2 =0.125M
Kc= [I2] [Cl2]/ [ICl]2 =0.11
1/0.11= [ICl]2/ [I2] [Cl2]
9.1 = [ICl]2 /[I2] [Cl2]
Let x= drop in concentration of I2 and Cl2 to reach equilibrium
At equilibrium
[I2] =0.125-x [CL2] =0.125-x [ICl] =2x
4x2/ (0.125-x)2= 9.1
Taking square root both sides, 2x/(0.125-x)= 3.02
2x = 3.02*(0.125-x) , 2x = 3.02*0.125- 3.02x, 5.02x = 3.02*0.125, x =0.0752
Equilibrium concentration [ICl] =2*0.0752=0.1504M, [I2] = [CL2] =0.125-0.0752 =0.0498M