Answer:
118.22 atm
Explanation:
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
KP = 0.13 = ![\frac{p(SO_{3})^{2}}{p(SO_{2})^{2}p(O_{2})}](https://tex.z-dn.net/?f=%5Cfrac%7Bp%28SO_%7B3%7D%29%5E%7B2%7D%7D%7Bp%28SO_%7B2%7D%29%5E%7B2%7Dp%28O_%7B2%7D%29%7D)
Where p(SO₃) is the partial pressure of SO₃, p(SO₂) is the partial pressure of SO₂ and p(O₂) is the partial pressure of O₂.
- With 2.00 mol SO₂ and 2.00 mol O₂ if there was a 100% yield of SO₃, then 2 moles of SO₃ would be produced and 1.00 mol of O₂ would remain.
- With a 71.0% yield, there are only 2*0.71 = 1.42 mol SO₃, the moles of SO₂ that didn't react would be 2 - 1.42 = 0.58; and the moles of O₂ that didn't react would be 2 - 1.42/2 = 1.29.
The total number of moles is 1.42 + 0.58 + 1.29 = 3.29. With that value we can calculate the molar fraction (X) of each component:
The partial pressure of each gas is equal to the total pressure (PT) multiplied by the molar fraction of each component.
Rewriting KP and solving for PT:
![\frac{p(SO_{3})^{2}}{p(SO_{2})^{2}p(O_{2})}=0.13\\\frac{(0.432*P_{T})^{2}}{(0.176*P_{T})^{2}(0.392*P_{T})} =0.13\\\frac{0.1866*P_{T}^{2}}{0.0121*P_{T}^{3}} =0.13\\\frac{15.369}{P_{T}}=0.13\\P_{T}=118.22 atm](https://tex.z-dn.net/?f=%5Cfrac%7Bp%28SO_%7B3%7D%29%5E%7B2%7D%7D%7Bp%28SO_%7B2%7D%29%5E%7B2%7Dp%28O_%7B2%7D%29%7D%3D0.13%5C%5C%5Cfrac%7B%280.432%2AP_%7BT%7D%29%5E%7B2%7D%7D%7B%280.176%2AP_%7BT%7D%29%5E%7B2%7D%280.392%2AP_%7BT%7D%29%7D%20%3D0.13%5C%5C%5Cfrac%7B0.1866%2AP_%7BT%7D%5E%7B2%7D%7D%7B0.0121%2AP_%7BT%7D%5E%7B3%7D%7D%20%3D0.13%5C%5C%5Cfrac%7B15.369%7D%7BP_%7BT%7D%7D%3D0.13%5C%5CP_%7BT%7D%3D118.22%20atm)
Answer:
is capable of combining with oxygen to form iron oxide
The enthalpy of vaporization of Bromine is 15.4 kJ/mol. -7.7 kJ is the energy change when 80.2 g of Br₂ condenses to a liquid at 59.5°C.
<h3>What is Enthalpy of Vaporization ?</h3>
The amount of enthalpy or energy that must be added to a liquid substance into gas substance is called Enthalpy of Vaporization. It is also known as Latent heat of vaporization.
<h3>How to find the energy change from enthalpy of vaporization ?</h3>
To calculate the energy use this expression:
![Q = n \Delta H_{\text{vapo.}](https://tex.z-dn.net/?f=Q%20%3D%20n%20%5CDelta%20H_%7B%5Ctext%7Bvapo.%7D)
where,
Q = Energy change
n = number of moles
= Molar enthalpy of vaporization
Now find the number of moles
Number of moles (n) = ![\frac{\text{Given Mass}}{\text{Molar mass}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BGiven%20Mass%7D%7D%7B%5Ctext%7BMolar%20mass%7D%7D)
= ![\frac{80.2\ g }{159.8\ g/mol}](https://tex.z-dn.net/?f=%5Cfrac%7B80.2%5C%20g%20%7D%7B159.8%5C%20g%2Fmol%7D)
= 0.5 mol
Now put the values in above formula we get
[Negative sign is used because Br₂ condensed here]
= - (0.5 mol × 15.4 kJ/mol)
= - 7.7 kJ
Thus from the above conclusion we can say that The enthalpy of vaporization of Bromine is 15.4 kJ/mol. -7.7 kJ is the energy change when 80.2 g of Br₂ condenses to a liquid at 59.5°C.
Learn more about the Enthalpy of Vaporization here: brainly.com/question/13776849
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Molecular Formula of Propanoic Acid is....
C₃H₆O₂Where;
Atomic Mass of C = 12
Atomic Mass of H = 1
Atomic Mass of O = 6
Substituting Atomic Masses in molecular formula,
(12)₃ + (1)₆ + (16)₂
36 + 6 + 32
74 g/mol
Result:
Molecular Mass of Propanoic Acid is 74 g/mol. Structure of Methyl Ethanoate and Propanoic Acid are as follow,
Answer:
The atoms are separating, and sublimation is happening.
Explanation: