An equilibrium mixture contains N2O4, (P= 0.30 ) and NO2 (P= 1.1 ) at 350 K. The volume of the container is doubled at constant temperature.
Calculate the equilibrium pressure of when the system reaches a new equilibrium.
Calculate the equilibrium pressure of NO2 when the system reaches a new equilibrium.
You don't give units for the pressure. Is that atm? torr? what?
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it's in atm
To calculate the equilibrium pressure of N2O4 when the system reaches a new equilibrium after the volume is doubled, we can use the ideal gas law:
PV = nRT
Where:
P is the pressure,
V is the volume,
n is the number of moles,
R is the ideal gas constant (0.0821 L·atm/mol·K), and
T is the temperature in Kelvin.
Let's define the initial and final conditions:
Initial: N2O4 (P = 0.30 atm) and NO2 (P = 1.1 atm)
Final: N2O4 (P') and NO2 (P'')
Given that the volume is doubled, we can say:
V' = 2V
Since the number of moles (n) remains constant, we can rewrite the ideal gas law as:
P'V' = nRT
To find the new equilibrium pressure of N2O4, we need to solve for P'.
1) Initial equation for N2O4:
PV = nRT
0.30 atm * V = nRT
2) Final equation for N2O4:
P'V' = nRT
P' * 2V = nRT
P' * (2 * 0.30 atm * V) = nRT
P' * 0.60 atm * V = nRT
Now, we can compare the initial and final equations:
0.30 atm * V = P' * 0.60 atm * V
Simplifying:
0.30 atm = 0.60 atm * P'
Now, we can solve for P':
P' = 0.30 atm / 0.60 atm
P' = 0.50 atm
Therefore, the equilibrium pressure of N2O4 when the system reaches a new equilibrium after the volume is doubled is 0.50 atm.
To calculate the equilibrium pressure of NO2 when the system reaches a new equilibrium, we need to consider the fact that the number of moles of NO2 will change due to the change in volume.
1) Initial equation for NO2:
PV = nRT
1.1 atm * V = nRT
2) Final equation for NO2:
P''V' = n''RT
P'' * 2V = n''RT
P'' * (2 * 0.30 atm * V) = n''RT
P'' * 0.60 atm * V = n''RT
Now, we can compare the initial and final equations:
1.1 atm * V = P'' * 0.60 atm * V
Simplifying:
1.1 atm = 0.60 atm * P''
Now, we can solve for P'':
P'' = 1.1 atm / 0.60 atm
P'' = 1.83 atm
Therefore, the equilibrium pressure of NO2 when the system reaches a new equilibrium after the volume is doubled is 1.83 atm.