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:
Kc = 0.20
Explanation:
N₂O₄ ⇄ 2NO₂
moles 5.3mol 2.3mol
Vol 5L 5L
Molarity 5.3/5M 2.3/5M
= 1.06M = 0.46M
Kc = [NO₂]²/[N₂O₄] = (0.46)²/(1.06) = 0.1996 ≅ 0.20
Answer:
Winnowing
Explanation:
Wind blows the lighter(in terms of mass) chaff from the whole grains,which are heavier(in terms of mass)
Nearer to the chlorine as it has a greater electronegativity
Answer:
Rate = 116m⁻¹s⁻¹[lactose][H]⁺
Explanation:
the formula for rate of reaction is given as
Rate = k[lactose]∧α[H]⁺∧β
we solve for the value of α and β
([lactose]₁/[lactose]₂)∧α
α = 
when we divide this equation
α = 
α = 1
we find β
R₁/R₂ = 0.01/0.02(0.001/0.001)∧β
0.00116/0.00232 = 0.5(1)∧β
β = 1
Rate = k[lactose]∧α[H]⁺∧β
we have to find the value for k
k = 0.00116/0.01(0.001)
k = 0.00116/0.00001
= 116m⁻¹s⁻¹
<u>Rate = 116m⁻¹s⁻¹[lactose][H]⁺</u>