Page 96 - Bruno Linder Elementary Physical Chemistry
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August 18, 2010 11:36 9in x 6in b985-ch07 Elementary Physical Chemistry
Chemical Kinetics 81
Solutions
a) Equilibrium Approximation:
2
2
2
k 1 [I 2 ]= k [I ] ; K =[I ] /[I 2 ]= k 1 /k ; [I ] =(k 1 /k )[I 2 ]
1
1
1
The rate-determining step is Step (2);
1
2
r = d[HI ]dt = k 2 [I ] [H 2 ]
2
d[HI ]/dt =2k 2 [I 2 ]= (2k 2 k 1 /k )[I 2 ][H 2 ]
1
b) Steady-State Approximation:
The intermediate substance, I, is
1
formed in Step (1) forward direction: d[I ]/dt = k 1[I 2 ]
2
1
reduced in Step (1) reverse direction: − d[I ]/dt = k [I ] 2
2 1
1
2
reduced in Step (2) − d[I ]/dt = k 2 [I ] [H 2]
2
Since the intermediate substance must disappear, we have
2
2
k 1 [I 2 ] − k [I ] − k 2[I ] [H 2]= 0
1
2
and so [I ] = k 1 [I 2]/(k + k 2 [H 2]). Thus, using Step 2 gives
1
d[HI ]/dt =2k 2[I 2 ][H 2 ]= 2k 2 k 1[I 2 ][H 2 ]/(k + k 2[H 2 ])
1
Realizing that Step (2) is the slow step, it means that k 2 k and so
1
neglecting the k 2 term in the denominator gives
d[HI ]/dt ≈ (2k 2 k 1 /k )[I 2][H 2]
1
which is the same as in Part (a).