Test Index

NCERT Class XII Chemistry
Chapter - Chemical Kinetics
Questions with Solutions

© examsnet.com
Question : 22
Total: 39
The rate constant for the decomposition of N2O5 at various temperatures is given below:
 T/°C  0  20  40  60  80
 105×k∕s−1  0.0787  1.70  25.7  178  2140
Draw a graph between ln k and 1/T and calculate the value of A and Ea.Predict the rate constant at 30°C and 50°C.
Solution:  
The values of rate constants for the decomposition of N2O5 at varioustemperatures are given below :
 T(°C)  T (K)  1/T  k (s‌–1)  ln k (= 2.303 log k)
 0  273  3.6 ×10‌−3  7.87 ×10‌−7  -14.06
 20  293  3.4 ×10‌−3  1.70×10‌−5  -10.98
 40  313  3.19 ×10‌−3  25.7 ×10‌−5  -8.266
 60  333  3.00 ×10‌−3  178×10‌−5  - 6.332
 80  353  2.8 ×10‌−3  2140 ×10‌−5  -3.844
Slope of the line = tan θ
=
y2−y1
x2−x1

=
−10.98−(−14.08)
3.4−3.6
×103
=−15.5×103
Ea=− slope ×R
=− (−15.5×103×8.314 )
=128.86 kJ K−1mol−1
‌Again‌ln‌A=ln‌k+
Ea
RT

=−14.06+
128.86×103JK−1mol−1
8.314×273

=−14.06+56.77=42.71
or, log‌A=18.53
or, A= antilog 18.53
=0.3388×1019
or,A=3.388×1018
Value of rate constant k at 303 K and 323 K can be obtained from graph.First of all ln k is obtained corresponding to
1
303
Kand
1
323
Kand thenk is calculated.
© examsnet.com
Go to Question:
Prev Question
Next Question