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Nitrosyl bromide, NOBr, is formed from NO and Br2.
2 NO(g) + Br2(g) 2 NOBr(g)
Experiments show the reaction is second order in NO and first order in Br2.
(a) What is the rate equation for the reaction?
rate = k[NO]2[Br2]
rate = k[NO][Br2]
rate = k[NO]2[Br2]2
rate = k[NO]2[Br]2

(b) How does the reaction rate change if the concentration of Br2 is changed from 0.0022 mol/L to 0.0066 mol/L?
the rate will double
the rate will remain unchanged
the rate will triple

(c) What is the change in the reaction rate if the concentration of NO is changed from 0.0024 mol/L to 0.0048 mol/L?
the rate will increase by a factor of 4
the rate will remain unchanged
the rate will also decrease by a factor of 1/2

Respuesta :

Answer:

(a) What is the rate equation for the reaction?

rate = k[NO]2[Br2]

Explanation:

(b) How does the reaction rate change if the concentration of Br2 is changed from 0.0022 mol/L to 0.0066 mol/L?

the rate will doubleNitrosyl bromide, NOBr, is formed from NO and Br2.

(a) Rate = [tex]k[NO]2[Br2][/tex]

(b) The rate will triple Nitrosyl bromide, NOBr, is formed from NO and Br2.

(c) The rate will triple increase by a factor of 4

Calculation of Nitrosyl bromide

(a) The reaction is second order in NO ---> [tex]rate , r = k[NO]^2[/tex]

the reaction is first order in Br2 --> rate ,[tex]r = k[Br2]^1[/tex]

Therefore, from the above two total rate equation is , rate , [tex]r = k [NO]^2[/tex][tex][Br2]^1[/tex]

--> [tex]r = k[NO]^2 [Br2][/tex]

(b)The rate law is rate , [tex]r = k[NO]^2 [Br2][/tex]

------> rate , r is directly proportional to [Br2]

----->[tex]r / r’ = [Br2] / [Br2]’[/tex]

r = initial rate

r’ = final rate

[Br2] = initial conc = 0.0022 mol/L

[Br2]’ = final conc = 0.0066 mol/L

Plug the values we get[tex]r/r’ = 0.0022 / 0.0066[/tex]

r / r’ = 0.33

----> r’ = r / 0.33 = 3r

Therefore, the rate will triple

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(c) The rate law is rate ,[tex]r = k[NO]^2 [Br2][/tex]

------> rate , r is directly proportional to [tex][NO]^2[/tex]

-----> r / r’ =[tex][NO]^2 / [NO]’^2[/tex]

r = initial rate[tex]r / r’ = [NO]^2 / [NO]’^2[/tex]

r’ = final rate

[NO] = initial conc = 0.0024 mol/L

[NO]’ = final conc = 0.0048 mol/L

Plug the values we get [tex]r/r’ = 0.0024 / 0.0048[/tex]

[tex]r / r’ = (0.0024^2) / (0.0048)^2[/tex]

= [tex](0.5)^2[/tex]

=[tex]0.25[/tex]

----> r’ = [tex]r / 0.25 = 4r[/tex]

Therefore, the rate will triple increase by a factor of 4

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