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 =
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 = ![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)
Hence, 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)
Learn more about rate law here:
brainly.com/question/4222261
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<span> Au</span>₂(SeO₄)₃
O = -2 × 4 = -8
Se = + 6
So,
(+6 - 8) = -2
Means (SeO₄) contains -2 charge, Now multiply -2 by 3
-2 ₓ 3 = -6
Means,
Au₂ + (-6) = 0
Au₂ = +6
Or,
Au = 6 / 2
Au = +3
Result:
Au = +3
Se = +6
O = -2
Ni(CN)₂
Cyanide (CN⁻) contains -1 charge,
So,
N = -3
C = +2
Then,
Ni + (-1)₂ = 0
Ni - 2 = 0
Or,
Ni = +2
Result:
N = -3
C = +2
Ni = +2
Mole = Mass / Molar mass
6.12 moles = Mass / 74.92 g/mol
Mass = 6.12 moles x 74.92 g/mol
Mass = 458.51g
1/8 got it from Yahoo answers and the answer was highlighted with 4 likes.... so most likely thats it
Answer:
The bronsted- Lowry acid is H₂PO₄⁻
Explanation:
Bronsted-Lowry acid donates a proton (H⁺)
H₂PO₄⁻ + OH⁻ → HPO₄²⁻ + H₂O
In the reaction above, H₂PO₄⁻ is donating the proton to OH⁻ resulting in H₂O and the deprotonated species. This makes it a bronsted-Lowry acid.