Chemical Kinetics
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4.30 The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume. SO2Cl2(g)→ SO2(g) + Cl2(g)
Calculate the rate of the reaction when total pressure is 0.65 atm.
4.30 The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume. SO2Cl2(g)→ SO2(g) + Cl2(g)
Calculate the rate of the reaction when total pressure is 0.65 atm.
4.30 When t = 0, the total partial pressure is P0 = 0.5 atm
When time t = t, the total partial pressure is Pt = P0 + p
P0-p = Pt-2p, but by the above equation, we know p = Pt-P0
Hence, P0-p = Pt-2 (Pt-P0)
Thus, P0-p = 2P0 – Pt
We know that time
t= 2.303/K log R0 / R
Where, k- rate constant
[R]° -Initi
Upvote 
4.29 For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.
Calculate the rate constant.
4.29 For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.
Calculate the rate constant.
4.29 When t = 0, the total partial pressure is P0 = 35.0 mm of Hg
When time t = t, the total partial pressure is Pt = P0 + p
P0-p = Pt-2p, but by the above equation, we know p = Pt-P0 Hence, P0-p = Pt-2 ( Pt-P0)
Thus, P0-p = 2P0 – Pt
We know that time
t= 2.303/K log R0 / R
Where, k- rate constant
[R]° -In
Upvote 4.28 A first order reaction takes 40 min for 30% decomposition. Calculate t1/2.
4.28 A first order reaction takes 40 min for 30% decomposition. Calculate t1/2.
4.28 Given:
Time t = 40 min
When 30% decomposition is undergone, 70% is the concentration.
We know, time taken
t= 2.303/K log R0 / R
Where, k- rate constant
[R]0 - initial concentration
[R] - concentration at time 't'
40 = 2.303/K log R0 / 0.7 R0
40 = 2.303/K log 1 / 0.7
40 / 0.1549 = 2.303 / k
Upvote 4.27 For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
4.27 For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
4.27 Let, initial concentration be [R]°
Concentration at 90% completion be (100-90)/100)* [R]°
? Concentration at 90% be 0.1 [R]°
Concentration at 99% completion be (100-99)/100)* [R]°? Concentration at 99% be 0.01 [R]°
we know time, t= 2.303/K log R0 / R
Time taken for 90% completion is
T90 =
Upvote 4.26 During nuclear explosion, one of the products is 90 Sr with half-life of 28.1 years. If 1µg of 90Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically?
4.26 During nuclear explosion, one of the products is 90 Sr with half-life of 28.1 years. If 1µg of 90Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically?
4.26 Initial concentration, [R]° = 1? g
Final concentration, [R] Half-life t1/2 = 28.1 years Solution:
We know, t1/2 = 0.693/k Where, k – rate constant
? k = 0.693/ t1/2
? k = 0.693/ (28.1 yrs)
? k = 0.0246 yrs-1
Also, t = 2.303 / k log R0 / R
If t = 10yrs, then, using the formula, we get,
t = 2.
Upvote 4.25 The rate constant for a first order reaction is 60 s–1. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?
4.25 The rate constant for a first order reaction is 60 s–1. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?
4.25 Given:
Order of the reaction = 1
Let, Initial concentration [R]°= x
Final concentration [R] = x/16
Rate constant k = 60 s-1
We know, time
t= 2.303 / k log R0 / R
t = 2.303 / 60 log (x/x/16)
t = 2.303 / 60s-1 log (1/1/16)
t = 2.303 X log 16 / 60s-1
Solving, we get t = 4.6 * 10-2s
Upvote 4.24 The experimental data for decomposition of N2O5 [2N2O5 → 4NO2 + O2 ] in gas phase at 318K are given below:
(i) Plot [N2O5 ] against t.
(ii) Find the half-life period for the reaction.
(iii) Draw a graph between log[N2O5 ] and t.
(iv) What is the rate law ?
(v) Calculate the rate constant.
(vi) Calculate the half-life period from k and compare it with (ii).
t/s
0
400
800
1200
1600
2000
2400
2800
3200
102 ×
[N2O5]/mol L-1
1.63
1.36
1.14
0.93
0.78
0.64
0.53
0.43
0.35
4.24 The experimental data for decomposition of N2O5 [2N2O5 → 4NO2 + O2 ] in gas phase at 318K are given below:
(i) Plot [N2O5 ] against t.
(ii) Find the half-life period for the reaction.
(iii) Draw a graph between log[N2O5 ] and t.
(iv) What is the rate law ?
(v) Calculate the rate constant.
(vi) Calculate the half-life period from k and compare it with (ii).
|
t/s |
0 |
400 |
800 |
1200 |
1600 |
2000 |
2400 |
2800 |
3200 |
|
102 × [N2O5]/mol L-1 |
1.63 |
1.36 |
1.14 |
0.93 |
0.78 |
0.64 |
0.53 |
0.43 |
0.35 |
4.24
(iv) As log [N2O5] vs time is a straight line given reaction is first Hence its rate law will be, Rate = k [N2O5]
(v) The slope of above graph is slope = 0.000209 K = 303 × slope
⇒4.82 × 10-4sec-1
Now, t1/2 = 0.693/K.
⇒0.693/4.82 × 10-4
⇒t1/2 = 1438 sec. which is almost equal to (ii)
Upvote <p><strong>4.24 The experimental data for decomposition of N<sub>2</sub>O<sub>5</sub> [2N<sub>2</sub>O<sub>5</sub> → 4NO<sub>2</sub> + O<sub>2</sub> ] in gas phase at 318K are given below: </strong></p>
<p><strong>(i) </strong>Plot [N2O5 ] against t.</p>
<p><strong>(ii)</strong> Find the half-life period for the reaction.</p>
<p><strong>(iii) </strong>Draw a graph between log[N2O5 ] and t.</p>
<p><strong> (iv) </strong>What is the rate law ?</p>
<p><strong> (v) </strong>Calculate the rate constant.</p>
<p><strong>(vi) </strong>Calculate the half-life period from k and compare it with (ii).</p>
4.23 The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample.
4.23 The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample.
4.23 Radio active decay occurs via first order rate law,
t1/2 = 5730 years. rate constant (k) of given decay is 0.693/t1/2
⇒0.693/5730 = 1.2 × 10-4 year-1
By first order integrated rate law age of the sample will be,
T = ( 2.303 / 1.2 X 10-4 ) log (A0/At)
where T is the age of the sample, A0 is t
Upvote