Answer:
Explanation:
From the given information:
A → B k₁
B → A k₂
B + C → D k₃
The rate law = ![\dfrac{d[D]}{dt}=k_3[B][C] --- (1)](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5BD%5D%7D%7Bdt%7D%3Dk_3%5BB%5D%5BC%5D%20---%20%281%29)
![\dfrac{d[B]}{dt}=k[A] -k_2[B] -k_3[B][C]](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5BB%5D%7D%7Bdt%7D%3Dk%5BA%5D%20-k_2%5BB%5D%20-k_3%5BB%5D%5BC%5D)
Using steady-state approximation;
![\dfrac{d[B]}{dt}=0](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5BB%5D%7D%7Bdt%7D%3D0)
![k_1[A]-k_2[B]-k_3[B][C] = 0](https://tex.z-dn.net/?f=k_1%5BA%5D-k_2%5BB%5D-k_3%5BB%5D%5BC%5D%20%3D%200)
![[B] = \dfrac{k_1[A]}{k_2+k_3[C]}](https://tex.z-dn.net/?f=%5BB%5D%20%3D%20%5Cdfrac%7Bk_1%5BA%5D%7D%7Bk_2%2Bk_3%5BC%5D%7D)
From equation (1), we have:
![\mathbf{\dfrac{d[D]}{dt}= \dfrac{k_3k_1[A][C]}{k_2+k_3[C]}}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cdfrac%7Bd%5BD%5D%7D%7Bdt%7D%3D%20%5Cdfrac%7Bk_3k_1%5BA%5D%5BC%5D%7D%7Bk_2%2Bk_3%5BC%5D%7D%7D)
when the pressure is high;
k₂ << k₃[C]
![\dfrac{d[D]}{dt} = \dfrac{k_3k_1[A][C]}{k_3[C]}= k_1A \ \ \text{first order}](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5BD%5D%7D%7Bdt%7D%20%3D%20%5Cdfrac%7Bk_3k_1%5BA%5D%5BC%5D%7D%7Bk_3%5BC%5D%7D%3D%20k_1A%20%5C%20%5C%20%20%5Ctext%7Bfirst%20order%7D)
k₂ >> k₃[C]
![\dfrac{d[D]}{dt} = \dfrac{k_3k_1[A][C]}{k_2}= \dfrac{k_1k_3}{k_2}[A][C] \ \ \text{second order}](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5BD%5D%7D%7Bdt%7D%20%3D%20%5Cdfrac%7Bk_3k_1%5BA%5D%5BC%5D%7D%7Bk_2%7D%3D%20%5Cdfrac%7Bk_1k_3%7D%7Bk_2%7D%5BA%5D%5BC%5D%20%5C%20%5C%20%20%5Ctext%7Bsecond%20order%7D)
Answer:
42686.04375
Explanation:
847.3219*34.6=2317.33774 multiply this by the next number 1.4560 to get 42686.04375
The freezing point depression is calculated through the equation,
ΔT = (kf) x m
where ΔT is the difference in temperature, kf is the freezing point depression constant (1.86°C/m), and m is the molality. Substituting the known values,
5.88 = (1.86)(m)
m is equal to 3.16m
Recall that molality is calculated through the equation,
molality = number of mols / kg of solvent
number of mols = (3.16)(1.25) = 3.95 moles
Then, we multiply the calculated amount in moles with the molar mass of ethylene glycol and the answer would be 244.9 g.
Melting, vaporization, boiling, and sublimation.