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Tardigrade
Question
Chemistry
For the reaction, N2 O5 (g) longrightarrow 2NO2 (g) +(1/2) O2 (g) The value of rate of disappearance of N2 O5 is given as 6.25 x 10-3 mol L-1 s-1. The rate of formation of NO2 and O2 is given respectively as
Q. For the reaction,
N
2
O
5
(
g
)
⟶
2
N
O
2
(
g
)
+
2
1
O
2
(
g
)
The value of rate of disappearance of
N
2
O
5
is given as 6.25 x
1
0
−
3
m
o
l
L
−
1
s
−
1
. The rate of formation of
N
O
2
and
O
2
is given respectively as
4019
211
AIPMT
AIPMT 2010
Chemical Kinetics
Report Error
A
6.25 x
1
0
−
3
m
o
l
L
−
1
s
−
1
an
d
6.25
×
1
0
−
3
m
o
l
L
−
1
s
−
1
8%
B
1.25 x
1
0
−
2
m
o
l
L
−
1
s
−
1
an
d
3.125
×
1
0
−
3
m
o
l
L
−
1
s
−
1
75%
C
6.25 x
1
0
−
3
m
o
l
L
−
1
s
−
1
an
d
3.125
×
1
0
−
3
m
o
l
L
−
1
s
−
1
11%
D
1.25 x
1
0
−
2
m
o
l
L
−
1
s
−
1
an
d
6.25
×
1
0
−
3
m
o
l
L
−
1
s
−
1
7%
Solution:
Key Idea Rate of disappearance of reactant = rate of appearance of product
or
−
Stoichiometric coefficient of reactant
1
d
t
d
[
reactant
]
=
+
Stoichiometric coefficient of product
1
d
t
d
[
product
]
For the reaction,
N
2
O
5
(
g
)
⟶
2
N
O
2
(
g
)
+
2
1
O
2
(
g
)
d
t
−
d
[
N
2
O
5
]
=
+
2
1
d
t
d
[
N
O
2
]
∴
d
t
d
[
N
O
2
]
=
−
2
d
t
d
[
N
2
O
5
]
=
2
×
6.25
×
1
0
−
3
m
o
l
L
−
1
s
−
1
=
12.5
×
1
0
−
3
m
o
l
L
−
1
s
−
1
=
1.25
×
1
0
−
2
m
o
l
L
−
1
s
−
1
d
t
d
[
O
2
]
=
−
d
t
d
[
N
2
O
5
]
×
2
1
=
2
6.25
×
1
0
−
3
m
o
l
L
−
1
s
−
1
=
3.125
×
1
0
−
3
m
o
l
L
−
1
s
−
1