Nitrogen gas (N2) reacts with hydrogen gas (H2) to form ammonia (NH3). At 200
oC in a closed
container, 1.1 atm of nitrogen gas is mixed with 2.1 atm of hydrogen gas in an otherwise empty container.
At equilibrium, the total pressure is 2.2 atm. Calculate the partial pressure of hydrogen gas at equilibrium.
Is there a Kp? Kc?
there isnt a kp or a kc
Try this.
..........N2 + 3H2 ==> 2NH3
initial...1.1..2.1......0
change...-x....-3x......2x
equil...1.1-x..2.1-3x...2x
Ptotal = 2.2 = 1.1-x+2.1-3x+2x
Solve for x and substitute into 2.1-3x for pH2. Something like 0.6 atm. for pH2.
Thanks it worked perfectly
To calculate the partial pressure of hydrogen gas at equilibrium, we need to use the concept of the mole ratios and the ideal gas law.
First, let's determine the moles of nitrogen gas and hydrogen gas present initially. We'll use the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
At 200 oC, we need to convert it to Kelvin by adding 273.15. So, the temperature T = (200 + 273.15) K = 473.15 K.
Using the ideal gas law, we can calculate the initial moles of nitrogen gas (n1) and hydrogen gas (n2):
n1 = (P1 * V) / (R * T) = (1.1 atm * V) / (R * 473.15 K)
n2 = (P2 * V) / (R * T) = (2.1 atm * V) / (R * 473.15 K)
Since the container is otherwise empty, the total moles of nitrogen gas and hydrogen gas at equilibrium must be equal to the sum of the initial moles:
n1 + n2 = n1' + n2'
Now, let's consider the balanced chemical equation for the reaction:
N2 + 3H2 -> 2NH3
According to the balanced equation, for every 3 moles of hydrogen gas (H2), 2 moles of ammonia (NH3) will be produced. Thus, the moles of nitrogen gas present at equilibrium will remain the same, but the moles of hydrogen gas will change:
n1' = n1
n2' = n2 - (2/3) * n2 (since the mole ratio is 3:2)
Finally, let's calculate the partial pressure of hydrogen gas at equilibrium. Since the total pressure at equilibrium is given as 2.2 atm, we can obtain the partial pressure of hydrogen gas (P2') using the ideal gas law:
P2' = (n2' * R * T) / V
= ((n2 - (2/3) * n2) * R * 473.15 K) / V
Substituting the initial pressure values for nitrogen gas (P1), hydrogen gas (P2), and the given volume (V), we can solve for P2':
P2' = ((2.1 atm - (2/3) * 2.1 atm) * R * 473.15 K) / V
= ((2.1 atm - 1.4 atm) * R * 473.15 K) / V
= (0.7 atm * R * 473.15 K) / V
So, the partial pressure of hydrogen gas (P2') at equilibrium is (0.7 atm * R * 473.15 K) / V.