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For the reaction 2H(2)(g)+2NO(g) rarr ...

For the reaction
`2H_(2)(g)+2NO(g) rarr N_(2)(g)+2H_(2)O(g)`
the observed rate expression is, rate `=k_(f)[NO]^(2)[H_(2)].` The rate experssion for the reverse reaction is :

A

`k_(b) [N_(2)][H_(2)O]^(2)//[H_(2)]`

B

`k_(b) [N_(2)][H_(2)O]`

C

`k_(b) [N_(2)][H_(2)O]^(2)//[NO]`

D

`k_(b) [N_(2)][H_(2)O]^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
`2H_(2) (g)+ 2NO(g) rarr N_(2)(g) +2H_(2)O(g)`
rate `=k_(f) [NO]^(2) [H_(2)]^(2) rArr K_(eq) =(k_(f))/(k_(b)) =([N_(2)][H_(2)O]^(2))/([NO]^(2) [H_(2)]^(2))`
At equilibrium, `r_(f) =r_(b)`
`k_(f) [H_(2)][NO]^(2) =(k_(b) [N_(2)][H_(2)O]^(2))/([H_(2)])`
Hence rate expression for reverse rxn
rate `=k_(b) ([N_(2)][H_(2)O]^(2))/([H_(2)])`
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