The temperature of a sample of gas in steel container at 30.3 Atm is
2 answers:
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Answer : The energy removed must be, 29.4 kJ
Explanation :
The process involved in this problem are :
The expression used will be:
where,
= heat released for the reaction = ?
m = mass of benzene = 94.4 g
= specific heat of solid benzene =
= specific heat of liquid benzene =
= enthalpy change for fusion =
Now put all the given values in the above expression, we get:
Negative sign indicates that the heat is removed from the system.
Therefore, the energy removed must be, 29.4 kJ
Answer:
N₂ = 0.7515atm
O₂ = 0.1715atm
NO = 0.0770atm
Explanation:
For the reaction:
N₂(g) + O₂(g) ⇄ 2NO(g)
Where Kp is defined as:
Pressures in equilibrium are:
N₂ = 0.790atm - X
O₂ = 0.210atm - X
NO = 2X
Replacing in Kp:
0.0460 = [2X]² / [0.790atm - X] [0.210atm - X]
0.0460 = 4X² / 0.1659 - X + X²
0.0460X² - 0.0460X + 7.6314x10⁻³ = 4X²
-3.954X² - 0.0460X + 7.6314x10⁻³ = 0
Solving for X:
X = - 0.050 → False answer. There is no negative concentrations.
X = <em>0.0385 atm</em> → Right answer.
Replacing for pressures in equilibrium:
N₂ = 0.790atm - X = <em>0.7515atm</em>
O₂ = 0.210atm - X = <em>0.1715atm</em>
NO = 2X = <em>0.0770atm</em>