D = m / V
D = 8.0 g / 25 cm³
D = 0.32 g/cm³
Answer A
The overall reaction is given by:

The fast step reaction is given as:

The slow step reaction is given as:
(slow step
)
Now, the expression for the rate of reaction of fast reaction is:
![r_{1}=k_{1}[NO][Br_{2}]-k_{-1}[NOBr_{2}]](https://tex.z-dn.net/?f=r_%7B1%7D%3Dk_%7B1%7D%5BNO%5D%5BBr_%7B2%7D%5D-k_%7B-1%7D%5BNOBr_%7B2%7D%5D)
The expression for the rate of reaction of slow reaction is:
Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as
takes place in this reaction.
The expression of rate of formation is:

=
(1)
Now, consider that the fast step is always is in equilibrium. Therefore, 
![k_{1}[NO][Br_{2}]= k_{-1}[NOBr_{2}]](https://tex.z-dn.net/?f=k_%7B1%7D%5BNO%5D%5BBr_%7B2%7D%5D%3D%20k_%7B-1%7D%5BNOBr_%7B2%7D%5D)
![[NOBr_{2}] = \frac{k_{1}}{k_{-1}}[NO][Br_{2}]](https://tex.z-dn.net/?f=%5BNOBr_%7B2%7D%5D%20%3D%20%5Cfrac%7Bk_%7B1%7D%7D%7Bk_%7B-1%7D%7D%5BNO%5D%5BBr_%7B2%7D%5D)
Substitute the value of
in equation (1), we get:
![\frac{d(NOBr)}{dt}=k_{2}[NOBr_{2}][NO]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%28NOBr%29%7D%7Bdt%7D%3Dk_%7B2%7D%5BNOBr_%7B2%7D%5D%5BNO%5D)
=![k_{2} \frac{k_{1}}{k_{-1}}[NO][Br_{2}][NO]](https://tex.z-dn.net/?f=k_%7B2%7D%20%5Cfrac%7Bk_%7B1%7D%7D%7Bk_%7B-1%7D%7D%5BNO%5D%5BBr_%7B2%7D%5D%5BNO%5D)
= ![\frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]](https://tex.z-dn.net/?f=%5Cfrac%7Bk_%7B1%7Dk_%7B2%7D%7D%7Bk_%7B-1%7D%7D%5BNO%5D%5E%7B2%7D%5BBr_%7B2%7D%5D)
Thus, rate law of formation of
in terms of reactants is given by
.
Answer:
The answer to your question is V2 = 333.9 ml
Explanation:
Data
Pressure 1 = P1 = 1.977 atm
Volume 1 = V1 = 517.4 ml
Pressure 2 = P2 = 3.063 atm
Volume 2 = V2 = ?
Process
To solve this problem use Boyle's law
P1V1 = P2V2
-Solve for V2
V2 = P1V1/P2
-Substitution
V2 = (1.977 x 517.4) / 3.063
-Simplification
V2 = 1022.9 / 3.063
-Result
V2 = 333.9 ml