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JEE Advanced Model Paper 14 with solutions for online practice
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© examsnet.com
Question : 36
Total: 60
For the reaction 2NO +
H
2
→
N
2
O
+
H_2O$$ , value of
(
−
d
p
d
t
)
changes from 1.5 torr
s
−
1
to 0.25 torr
s
−
1
when pressure is changed from 360 torr to 150 torr with respect to NO keeping pressure of hydrogen to be constant. If pressure of NO is kept constant, it is observed thatvalue of
(
−
d
p
d
t
)
changes from 1.6 torr
s
−
1
to 0.8 torr
s
−
1
with respect to pressure of hydrogen from 300 torr to 150 torr. Order of the reaction will be [log 2 = 0.301 ; log 3 = 0.477 ; log 5 = 0.699 ; log 7 = 0.845]
Your Answer:
Validate
Solution:
Let order with respect to NO be “x” and that of
H
2
be ’y'
r =
(
−
d
p
d
t
)
;
1.5
0.25
=
(
360
150
)
x
⇒ x = 2 ;
1.6
0.8
=
(
300
150
)
y
⇒ y = 1
Over all order of reaction = 3
© examsnet.com
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