Home
Class 12
CHEMISTRY
Consider the reaction : 2H(2)(g)+2NO(g...

Consider the reaction :
`2H_(2)(g)+2NO(g)rarrN_(2)(g)+2H_(2)(g)`
The rate law for this reaction is :
Rate = `k[H_(2)][NO]^(2)`
Under what conditions could these steps represent mechanism?
`{:("Step 1:",2NO(g)hArrN_(2)O_(2)(g)),("Step 2:",N_(2)O_(2)+H_(2)rarrN_(2)O+H_(2)O),("Step 3:",N_(2)O+H_(2)rarrH_(2)O+N_(2)):}`

A

Step 1 is the slow step

B

Step 2 is the slow step

C

Step 3 is the slow step

D

Step 1 and 2 are slow steps

Text Solution

Verified by Experts

The correct Answer is:
B
`r=k[N_(2)O_(2)][H_(2)]=k_(eq) ([N_(2)O_(2)])/([NO]^(2))" "r=k K_(eq) =[NO]^(2)[H_(2)]=k[NO]^(2)`
Promotional Banner

Similar Questions

Explore conceptually related problems

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 :

For the reaction , 2C_(2)H_(6)(g)+7O_(2)(g)rarr4CO_(2)(g)+6H_(2)O(l) , the rate of disappearnce of C_(2)H_(6)(g) :

2H_(2)(g)+O_(2)(g)to2H_(2)O(l)+xkJ In the above reaction

For the reaction 2NO(g)+2H_(2)(g)toN_(2)(g)+2H_(2)O(g) . The rate law is rate =k[NO]^(2)[H_(2)] . What is the overall order of reaction.

The reaction 2H_2O(l)rarr2H_2(g)+O_2(g) is an example of

consider the reaction . 2H_(2)(g)+2NO(g)to N_(2)(g)+2H_(2)O(g) The rate low of this reaction will be ,if mechanism of reaction is Step1: 2NOhArrN_(2)O_(2)["fast equilibrium"] Step2: N_(2)O_(2)+H_(2)toN_(2)O+H_(2)O["slow"] Step3: N_(2)O+H_(2)toN_(2)+H_(2)O["fast"]