What is the reaction order at each temperature? The following data were obtained
ID: 481004 • Letter: W
Question
What is the reaction order at each temperature? The following data were obtained for the decomposi- tion of nitrogen dioxide. 2NO2(g) 2NO(g) O2 (g) Pressure of NO2 (torr) Time (s) 310K 315 K 24.0 24.0 15.2 18.1 13.7 9.7 6.1 10.3 7.8 3.9 2.5 5.9 1.6 4.5 3.4 1.0 0.6 1.9 0.4 10 1.5 0.3 Select one or more a. Reaction order at 310 K: 0 b. Reaction order at 310 K: 1 c. Reaction order at 310 K: 2 d. Reaction order at 315 K: 0 e. Reaction order at 315 K: 1 f Reaction order at 315 K: 2 What is the rate constant at 310 K? (Use Sl units. For example, use seconds not minutes or hours.) Answer: What is the rate constant at 315K? Answer: What is the activation energy (in J/mol)? Answer:Explanation / Answer
The given reaction is
2 NO2 (g) -------> 2 NO (g) + O2 (g)
The rate law for the reaction is
-dP/dt = kPn where P is the pressure of NO2 in torr. We need to evaluate n and K.
Plot a table of ln P vs t at the two different temperatures:
Time (s)
Pressure (P) of NO2 at 310 K (torr)
ln P at 310 K
Pressure (P) of NO2 at 315 K (torr)
ln P at 315 K
0
24.0
3.178
24.0
3.178
1
18.1
2.896
15.2
2.721
2
13.7
2.617
9.7
2.272
3
10.3
2.332
6.1
1.808
4
7.8
2.054
3.9
1.361
5
5.9
1.775
2.5
0.916
6
4.5
1.504
1.6
0.470
7
3.4
1.224
1.0
0.000
8
2.6
0.955
0.6
-0.511
9
1.9
0.642
0.4
-0.916
10
1.5
0.405
0.3
-1.204
Plot of ln P vs t; Series 1 denotes the data at 310 K and series 2 denotes the data at 315 K.
Rate constant at 310 K
The reaction is first order in P (NO2), i.e, the rate law can be written as
-dP/dt = kP
===> dP/P = - k.dt
Integrate both sides to write
ln P = -kt + c where c = constant of integration ……(1)
The equation is of the form y = mx + c where m = slope of the linear plot. Comparing the two equations, we have,
Slope = m = -k
Now go back to the plot above; Slope = -0.2785 (Series 1, 310 K).
Therefore, -0.2785 = -k
===> k = 0.2785
We have obtained a numerical value for k, but we need to find out the unit now. Go back to equation (1) above. The left hand side is dimensionless; hence the right must also be dimensionless. t has the unit of s, hence k must have unit s-1.
Therefore, the rate constant at 310 K is 0.2785 s-1 (ans).
Rate constant at 315 K
Use the plot again (Series 2, 315 K).
Slope = -0.4483; hence -0.4483 = -k
===> k = 0.4483
As before the unit of k will be s-1.
The first order rate constant at 315 K is 0.4483 s-1 (ans).
Activation energy
We know, from the Arrhenius equation,
k = A.e^(-Ea/RT) where Ea is the activation energy of the reaction (constant for a particular reaction) and A is the Arrhenius constant.
Therefore,
ln k = ln A – Ea/RT
If we have two separate rate constants (k1 and k2) at two different temperature (T1 and T2), we must have,
ln k1 = ln A – Ea/RT1
ln k2 = ln A – Ea/RT2
Therefore,
ln k2 – ln k1 = ln A – Ea/RT2 – ln A + Ea/RT1
===> ln (k2/k1) = Ea/R*(1/T1 – 1/T2)
Plug in values from above
ln (0.4483 s-1/0.2785 s-1) = Ea/R*(1/310 – 1/315) K-1
===> ln (1.6097) = Ea/R*(5.1203*10-5) K-1
===> 0.476 = Ea/R*(5.1203*10-5 K-1)
===> Ea = (0.476)*(8.314 J/mol.K)/(5.1203*10-5 K-1)
===> Ea = 7.7289*104 J/mol 7.73*104 J/mol (ans).
Time (s)
Pressure (P) of NO2 at 310 K (torr)
ln P at 310 K
Pressure (P) of NO2 at 315 K (torr)
ln P at 315 K
0
24.0
3.178
24.0
3.178
1
18.1
2.896
15.2
2.721
2
13.7
2.617
9.7
2.272
3
10.3
2.332
6.1
1.808
4
7.8
2.054
3.9
1.361
5
5.9
1.775
2.5
0.916
6
4.5
1.504
1.6
0.470
7
3.4
1.224
1.0
0.000
8
2.6
0.955
0.6
-0.511
9
1.9
0.642
0.4
-0.916
10
1.5
0.405
0.3
-1.204
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