Answer: In metallic bonds, the mobile electrons surrounding the positive ions are called <u><em>dipole</em></u>.
The overall rate law in terms of the initial rate of the reaction is:
= ![k_{overall} [H_{2} SO_{4} ]^{2} [C_{6}H_{6}]](https://tex.z-dn.net/?f=k_%7Boverall%7D%20%5BH_%7B2%7D%20SO_%7B4%7D%20%5D%5E%7B2%7D%20%5BC_%7B6%7DH_%7B6%7D%5D)
The Sulfonation of benzene has the following mechanism:
(1) 
⇌ 
[fast]
(2) 
→ 
[slow]
(3) 
→ 
[fast]
(4) 
→ 
[fast]
<h3>Calculation of rate law:</h3>
Given that the reaction's slowest step is the rate-dependent step, the following is the definition of rate law:
Rate = ![k_{2} [SO_{3} ][C_{6} H_{6} ]](https://tex.z-dn.net/?f=k_%7B2%7D%20%5BSO_%7B3%7D%20%5D%5BC_%7B6%7D%20H_%7B6%7D%20%5D)
However, the issue is that because 
is an intermediate, it cannot be accounted for in the general rate legislation.
The following rate rule governs the synthesis of :
Rate = ![k_{1}[H_{2} SO_{4} ]^{2}](https://tex.z-dn.net/?f=k_%7B1%7D%5BH_%7B2%7D%20SO_%7B4%7D%20%20%5D%5E%7B2%7D)
Consequently, if we replace equation 1 with equation 2,
Rate is determined to be =
Hence, the overall rate law in terms of the initial rate of the reaction is:
=
Learn more about rate law here:
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Answer:
Na2S
Explanation:
Na2S: Na (always +1). S (-2) in this formula
CaSO4: Ca (always +2). O (-2). => S (+6) in CaSO4
S (0)
SO2: O(-2) => S (+4)
Hope this answer will help you! Have a nice day!
The rate of reaction is always expressed in concentration per time like mol/L·s. The equation is:
r [mol/L·s] = kCⁿ, where n is the order of reaction. Since k is 1300/s, that means that Cⁿ = C such that (1/s)*(mol/L) = mol/L·s. Thus, n=1. For a first order reaction, the formula would be:
ln(A/A₀) = -kt
where
A is the amount of material after time t
A₀ is the amount of material at t=0
The half life is when A/A₀ = 1/2÷1 = 1/2. Thus, the half-life t is:
ln(1/2) = (-1300t)
t = 5.33×10⁻⁴ seconds
