Billones Solutions
CHEM 18 SOLVED PROBLEMS 1. The gas-phase decomposition of NO2, NO2(g) the following data: Time (s) [NO2] (M)
0 0.100
5 0.017
20 0.0090
NO(g) + O2(g), is studied at 383 oC, giving 15 0.0062
20 0.0047
a) What is the order of reaction with respect to NO 2? b) What is the value of the rate constant? Solution:
a) n = 0 n=1
r = -0.779 r = -0.924
n=2 n=3
r = 0.999 r = 0.937
Since the dataset fits the plot for n = 2 (most linear, r ≈ 1), order = 2.
The reaction is SECOND order with respect to NO 2. b) Since the plot plot of ln[NO2] vs time is linear ( r ≈ 1), the reaction is second second order. The integrated rate equation for second order is 1 [ NO2 ]
=
kt +
1 [ NO2 ]0
ln
Evidently, the slope, m, is equal to the rate constant, k. The plot of ln[NO2] vs time has has a slope slope of 10. Thus, k = 10 M-1s-1 2. The following gas-phase reaction was studied at 290°C by observing the change in pressure as a function of time in a constant-volume vessel: ClCO2CCl3( g) g)
2COCl2( g) g)
Determine the order of the reaction and the rate constant based on the following data: Time (s) P of ClCO2CCl3 (mm Hg)
0 15.76
181 12.64
513 8.73
1164 4.44
Solution:
n=0 n=1
r = -0.979 r = -0.999
n=2 n=3
r = 0.991 r = 0.968
Since the dataset fits the plot for n = 1 (most linear, r ≈ -1), order = 1.
The reaction is FIRST order with respect to ClCO2CCl3. 3. Variation of the rate constant with temperature for the first-order reaction 2N2O5( g) g)
2N2O4( g) g) + O2( g) g)
is given in the following table. T (oC) k (s k (s-1)
25 -5 1.74 x 10
35 -5 6.61 x 10
45 -4 2.51 x 10
55 -4 7.59 x 10
a) Determine graphically the activation energy for the reaction. b) Determine the rate constant, k, and frequency factor, A, at 40 oC?
65 -3 2.40 x 10
Billones Solutions
Solution:
a) The slope of the plot of ln k vs (1/T in K) is equal to –Ea/R. The plot of ln k vs (1/T in K) has a slope, m, of -12392. The Ea = -mR = -(-12392 K) (8.314 x 10 -3 kJ/mol K) = 103 kJ/mol b) The plot of ln k vs (1/T in K) has a slope, m, of -12392 and y-intercept of 30.629. The equation of the straight line is ⎛1⎞ − E a ⎛ 1 ⎞ ln k = ⎜ ⎟ + ln A or ln k = (−12392)⎜ ⎟ + 30.629 R
⎝ T ⎠
⎝ T ⎠
o
At 40 C or 313 K,
⎛ 1 ⎞ ⎟ + 30.629 ⎝ 313 ⎠
ln k = ( −12392)⎜ ln k = −8.962 k = e
−8.962
=
−4 −1
1.28 x 10 s
Thus, the rate constant at 40 oC or 313 K is 1.28 x 10-4 s-1. b
The y-intercept, b, is equal to ln A. Thus, the frequency factor, A, is e . The y-intercept is 30.629; 30.629
A=e
13
= 2.0 x 10
4. Given the same reactant concentrations, the reaction CO( g) + Cl2( g)
COCl2( g)
at 250°C is 1.50 x 10 3 times as fast as the same reaction at 150°C. Calculate the activation energy for this reaction. Assume that the frequency factor is constant. Solution:
Since the rate at 250 oC (523 K) is 1.50 x 103 times as fast as that at 150 oC (423 K), at same concentration, 3 3 k 523 = 1.50 x 10 k 423 ; k 523 /k 423 = 1.50 x 10
⎛ T 523 − T 423 ⎞ E a ⎛ T 523 − T 423 ⎞ 3 x ⇒ ln(1.50 10 ) ⎜ ⎟ ⎜ ⎟ k 423 R ⎝ T 523T 423 ⎠ R ⎝ T 523T 423 ⎠ ⎡ 523K − 423K ⎤ E a 3 ln(1.50 x 10 ) ⎢ ⎥ −3 8.314 x 10 kJ /( mol.K ) ⎣ (523K )(423K ) ⎦ ln
k 523
E a
=
=
=
7.31322
E a =
−3
8.314 x 10 kJ /( mol.K )
(4.52 x 10
−4
/ K )
Ea = 135 kJ/mol 5. The bromination of acetone is acid-catalyzed: H+
CH3COCH2Br + H+ + Br_
CH3COCH3 + Br2 catalyst
The rate of disappearance of bromine was measured for several different concentrations of acetone, bromine, and H + ions at a certain temperature:
Billones Solutions
Run 1 2 3 4 5
[CH3COCH3] 0.30 0.30 0.30 0.40 0.40
[Br2] 0.050 0.100 0.050 0.050 0.050
Rate of Disappearance of Br2 (M/s) 5.7 x 10-5 5.7 x 10-5 1.2 x 10-4 3.1 x 10-4 7.6 x 10-5
+
[H ] 0.050 0.050 0.100 0.200 0.050
(a) What is the rate law for the reaction? (b) Determine the rate constant. (c) The following mechanism has been proposed for the reaction:
Show that the rate law deduced from the mechanism is consistent with that shown in (a). Solution:
a) In runs 1 and 2, the concentration of Br 2 was varied at constant [CH 3COCH3] and [H+]. Since the rate did not change the order with respect to Br 2 is zero. You can ignore the column on [Br 2] in evaluating the order with respect to other reactants. In runs 2 and 3, the concentration of H + was doubled (from 0.05 M to 0.10 M) at constant [CH3COCH3] and the rate of the reaction was doubled (from 0.000057 M to 0.00012 M). Thus, the reaction is first order in H+. In runs 1 and 4, the concentration of CH 3COCH3 was increased by 1 and 1/3 (0.30 M to 0.40 M) while keeping the concentration of H + constant. The rate of the reaction also increased by 1 and 1/3 (7.6 x 10 -5/5.7 x 10-5). The reaction is first order in CH3COCH3. Therefore, the rate law is given by Rate = k[CH3COCH3][H+] b) Using the data in run 1, the rate constant is k =
rate1 +
[CH 3COCH 3 ]1[ H ]1
=
5.7 x 10
−5
(0.30)(0.05)
k = 0.0038 M-1s-1 b) Since step 2 is the rate-determining step (rds), rate = k2[CH3C(=OH+)CH3][H2O] Applying the equilibrium principle to express the intermediate concentration [CH 3C(=OH+)CH3]
Billones Solutions
in terms of reactant concentration, we have +
K 1
[CH 3C ( = OH )CH 3 ][ H 2O]
=
+
[CH 3COCH 3 ][ H 3O ] +
+
[CH 3C ( = OH )CH 3 ] = K 1
[CH 3COCH 3 ][ H 3O ] [ H 2O]
⎧ [CH 3COCH 3 ][ H 3O rate = k 2 ⎨K 1 [ H 2O] ⎩ rate
=
k 2K 1[CH 3COCH 3 ][ H 3O
+
rate
=
k obs[CH 3COCH 3 ][ H 3O
+
]
+
]⎫ ⎬[ H 2O]
⎭
]
The derived rate law: Rate = k obs[CH3COCH3][H3O+] is consistent with the observed one (in a). 5. The thermal decomposition of phosphine (PH 3) into phosphorus and molecular hydrogen is a first-order reaction: 4PH3( g)
P4( g) + 6H2( g)
The half-life of the reaction is 35.0 s at 680°C. Calculate (a) the first-order rate constant for the reaction, (b) the time required for 95 percent of the phosphine to decompose, (c) the total pressure of the system after 100 s if the initial pressure is 150 mmHg of PH 3. Solution:
a) The rate constant is
ln 2
k
=
0.693 =
t 1/ 2
35 s
=
0.0198 s
−1
b) If 95% of PH3 has decomposed, [PH 3]0 = 1 and [PH3] = 1 – 0.95 = 0.05 ln
[ PH 3 ] [ PH 3 ]0
⎛ 0.05 ⎞ ⎟ ⎝ 1 ⎠
ln⎜
−2.9957 t
=
−kt
=
=
=
−(0.0198 s−1) t −(0.0198 s−1 )t
151 s
c) The final pressure of PH 3 after 100 s is calculated as follows. PPH ln − kt P0,PH 3
=
3
ln
ln
PPH
3
150 mm Hg
150 mm Hg
PPH
3
150 mm Hg
PPH
3
=
=
−1
−
=
−1.98
PPH
3
(0.0198s )100s
=
e −1.98
=
20.71 mm Hg
0.138
Billones Solutions
of PH3 150 -4x 150 – 4x
of P4 0 +x x
P
Initial Change Final
of H2 0 +6x 6x
P
P
The change in pressure of PH 3 is -4x = 20.71 – 150 = -129.29. Thus, x = 32.32 of PH3 150 – 4x 150 – 129.29 20.71
of P4 x 32.32 32.32
P
Final
of H2 6x 6(32.32) 193.94
P
P
The total pressure after 100 s is the sum of the partial pressures: Total pressure = 20.71 + 32.32 + 193.94 = 246.97 mm Hg 6. The rate law for the reaction 2H2( g) + 2NO( g)
N2( g) + 2H2O( g)
is rate = k obs[H2][NO]2. Propose at least two valid mechanisms for this reaction. Solution:
Mechanism A Step 1 Step 2
H2 + 2NO N2O + H2
N2O + H2O (slow, rds) N2 + H2O (fast)
Overall
2H2 + 2NO → N2 + 2H2O
→ →
The rate law based on step 1 is Rate = k 1[H2][NO]2
which is consistent with observed rate law.
Mechanism B Step 1 Step 2 Step 3
2NO ⇔ N2O2 N2O2 + H2 → N2O + H2O N2O + H2 → N2 + H2O
Overall
(fast, equilibrium) (slow, rds) (fast)
2H2 + 2NO → N2 + 2H2O
The rate law based on step 2 is Rate = k 2[N2O2][H2] Applying equilibrium principle in expressing [N 2O2] in terms of reactant concentration, we have K 1
=
[ N 2O2 ] [ NO]
(
2
[ N 2O2 ]
⇒ 2
)
rate
=
k 2 K 1[ NO] [ H 2 ]
rate
=
k obs[ H 2 ][ NO]
2
=
2
K 1[ NO]
Billones Solutions
N2O2 is formed in the forward reaction in step 1. However, it is consumed in the reverse reaction in step 1 and the forward reaction in step 2. Applying steady-state principle in expressing [N 2O2] in terms of reactant concentration, we have rate of formation of N2O2 = rate of consumption of N 2O2 rate1 = rate1’ + rate2 2
k 1[ NO]
2
k 1[ NO]
=
k 1 '[ N 2O2 ] + k 2 [ N 2O2 ][ H 2 ]
=
[ N 2O2 ]( k 1' + k 2 [ H 2 ]) 2
[ N 2O2 ] =
k 1[ NO]
(k ' 1
+ k 2
[ H 2 ])
Since step 1 is a lot faster than step 2, then k 1’ >>> k2, so that k1’+ k2[H2] = k1’. Thus, 2
[ N 2O2 ]
=
k 1[ NO] k 1 '
Substitution to the rate law based on step 2, we have rate
=
rate
=
rate
=
k 2 [ N 2O2 ][ H 2 ]
⎛ k 1[ NO]2 ⎞ k 2 ⎜ ⎟[ H 2 ] ⎝ k 1 ' ⎠ 2
k obs[ H 2 ][ NO]