A one liter flask with the two species at equilibrium contains 0.170 mole isobutane and 0.0680 mole n-butane. Then the concentra
1 answer:
Given question is incomplete. The complete question is as follows.
n-butane and isobutane are in equilibrium at 25 degrees centigrade.
n-butane(g) isobutane(g)
A one liter flask with the two species at equilibrium contains 0.170 mole isobutane and 0.0680 mole n-butane. Then the concentration of n-butane is increase by 0.200 moles. (0.200 mole n-butane is added.) What is the new equilibrium concentration for isobutane.
Explanation:
As the given reaction is as follows.
n-butane(g) isobutane(g)
Initial: x 0
Equilbm: x - 0.17 0.17
It is given that equilibrium concentration of n-butane is 0.068 M.
So, x - 0.17 = 0.0680
x = 0.238
Therefore, the initial concentration is 0.238.
Here,
=
= 2.5
When 0.2 mol of n-butane is added. Hence, total moles will be as follows.
Total mole = (0.238 + 0.2)
= 0.438 mol
Hence, for 1 L volume the ICE table will be as follows.
n-butane(g) isobutane(g)
Initial: 0.438 0
Equilbm: 0.438 - x x
2.5 =
1.095 - 2.5x = x
x =
= 0.3128 M
Thus, we can conclude that the equilibrium concentration is 0.3128 M.
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Answer:
The new temperature of the nitrogen gas is 516.8 K or 243.8 C.
Explanation:
Gay-Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of collisions with the walls increases, that is, the pressure increases. That is, the pressure of the gas is directly proportional to its temperature.
Gay-Lussac's law can be expressed mathematically as follows:
Where P = pressure, T = temperature, K = Constant
You want to study two different states, an initial state and a final state. You have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment. By varying the temperature to a new value T2, then the pressure will change to P2, and the following will be fulfilled:
In this case:
- P1= 2 atm
- T1= 50 C= 323 K (being 0 C= 273 K)
- P2= 3.2 atm
- T2= ?
Replacing:
Solving:
T2= 516.8 K= 243.8 C
<u><em>The new temperature of the nitrogen gas is 516.8 K or 243.8 C.</em></u>