following data are obtained for the reaction : N2O5 -----> 2NO2 + 1/2 O2 t/s 0 300 600 N2O5 (mol L^-1) 1.6×10^-2 0.8×10^-2 0.4×10^-2
Answers
Following data are obtained for the reaction :
N
2
O
5
→2NO
2
+
2
1
O
2
t/s 0 300 600
[N
2
O
5
]/molL
−1
1.6×10
−2
0.8×10
−2
0.4×10
−2
i. Show that it follows first order reactions .
ii. Calculate the half-life .
(Given log2=0.3010,log4=0.6021 )
Medium
Solution
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i. At 300s ,
k=
t
2.303
log
[R]
[R]
0
=
300
2.303
log
0.8×10
−2
1.6×10
−2
=
300
2.303
log2
k=
300
2.303
×0.3010=2.31×10
−3
s
−1
At 600s
k=
t
2.303
log
[R]
[R]
0
=
600
2.303
log
0.4×10
−2
1.6×10
−2
=
600
2.303
log4
k=
600
2.303
×0.6021=2.31×10
−3
s
−1
k is constant and is equal to 2.31×10
−3
s
−1
when we use first order equation. Hence , it follows first order reaction.
ii. t
1/2
=
k
0.603
=
2.31×10
−3
s
−1
0.693
=300s
Answer:
i. At 300s ,
k=t2.303log[R][R]0
=3002.303log0.8×10−21.6×10−2=3002.303log2
k=3002.303×0.3010=2.31×10−3s−1
At 600s
k=t2.303log[R][R]0
=6002.303log0.4×10−21.6×10−2=6002.303log4
k=6002.303×0.6021=2.31×10