Rate of formation of NO for the following reaction
2NOB → 2NO+ Br, is reported as 1.6 X10-4 molL-'s! The
rate of the reaction is found to be
A. 8 X10-5 molL-'s-1
B.3.2 X10-4 molL-'s-1
C.1.6 X104 moll-'s-1
D.4 X10-5 moll's!
Answers
Answered by
6
From the slow step, rate=k[NOBr
2
][NO] ...... (1)
From the equilibrium reaction, equilibrium constant K=
[NO][Br
2
]
[NOBr
2
]
[NOBr
2
]=K[NO][Br
2
]......(2)
Substitute equation (2) in equation (1).
rate=kK[NO][Br
2
][NO]=K
′
[NO]
2
[Br
2
]
❥︎hope it helps u dear
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