Part A Cyclohexane, C 6 H 12 , undergoes a molecular rearrangement in the presen

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

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