In the reaction of sodium and oxygen, which atom is the reducing agent?
1 answer:
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Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄
Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.
For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:
Kp =
where:
P(N₂O₄) and P(NO₂) are the partial pressure of each gas.
Calculating constant:
Kp =
Kp = 0.0104
After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.
P(N₂O₄) + P(NO₂) = 200
P(N₂O₄) = 200 - P(NO₂)
Kp =
0.0104 =
0.0104 +
- 200 = 0
Resolving the second degree equation:
=
= 98.7
Find partial pressure of N₂O₄:
P(N₂O₄) = 200 - P(NO₂)
P(N₂O₄) = 200 - 98.7
P(N₂O₄) = 101.3
The partial pressures are = 98.7 MPa and P(N₂O₄) = 101.3 MPa
Answer:
D. 15.8atm
Explanation:
Given parameters:
Initial pressure = 13atm
Initial temperature = 34°C = 34 + 273 = 307K
Final temperature = 100°C = 100 + 273 = 373K
Unknown:
Final pressure = ?
Solution:
To solve this problem, we apply a derivation of the combined gas law taking the volume as a constant.
The expression is shown mathematically below;
=
P and T pressure and temperature values
1 and 2 are initial and final states
Insert the parameters and solve for T₂;
=
P₂ = 15.8atm