a 20.0-L vessel at 700 K initially contains HI (g) at a pressure of 6.20 atm;at equilibrium, it is found that the partial pressure of H2 (g) is .600 atm. What is the partial pressure of HI (g) at equilibrium?
5.00 atm
Do you have a Kp or Kc?
To find the partial pressure of HI at equilibrium, we can use the balanced chemical equation for the reaction:
2 HI (g) ⇌ H2 (g) + I2 (g)
The equilibrium constant expression for this reaction is given by:
Kc = (PH2 × PI2) / P(HI)^2
We are given the initial pressure of HI (P(HI)) as 6.20 atm and the equilibrium partial pressure of H2 (PH2) as 0.600 atm. We need to find the partial pressure of HI at equilibrium.
Let's substitute the values into the equilibrium constant expression and solve for P(HI):
Kc = (0.600 × PI2) / (6.20)^2
To find the value of PI2, we need to know the equilibrium constant (Kc) for the reaction. Unfortunately, the equilibrium constant is not provided in the question. Without the value of Kc, it is not possible to determine the partial pressure of HI at equilibrium.
To find the partial pressure of HI gas at equilibrium, we can use the ideal gas law and the stoichiometry of the reaction between HI and H2. The balanced equation for the reaction is:
2HI (g) ↔ H2 (g) + I2 (g)
From the reaction equation, we can see that 2 moles of HI produce 1 mole of H2. So, the mole ratio of HI to H2 is 2:1.
Step 1: Calculate the number of moles of H2 at equilibrium:
Using the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Since we are given the initial pressure, we can rearrange the equation to solve for the number of moles:
n = PV / RT
Given:
Initial pressure of H2, P(H2) = 6.20 atm
Temperature, T = 700 K
Volume, V = 20.0 L
R = 0.0821 L·atm/(mol·K)
Plug in the values:
n(H2) = (6.20 atm) * (20.0 L) / (0.0821 L·atm/(mol·K) * 700 K)
n(H2) = 9.493 mol (rounded to the appropriate number of significant figures)
Step 2: Calculate the number of moles of HI at equilibrium:
Since we know the stoichiometry of the reaction, for every 1 mole of H2, there are 2 moles of HI. Therefore:
n(HI) = 2 * n(H2)
n(HI) = 2 * 9.493 mol
n(HI) = 18.986 mol (rounded to the appropriate number of significant figures)
Step 3: Calculate the partial pressure of HI at equilibrium:
Using the ideal gas law equation, we can now determine the partial pressure of HI:
P(HI) = n(HI) * RT / V
Where:
P(HI) = partial pressure of HI
n(HI) = number of moles of HI
T = temperature
R = ideal gas constant
V = volume
Given:
n(HI) = 18.986 mol
T = 700 K
V = 20.0 L
R = 0.0821 L·atm/(mol·K)
Plug in the values:
P(HI) = (18.986 mol) * (0.0821 L·atm/(mol·K)) * (700 K) / (20.0 L)
P(HI) ≈ 6.46 atm
Therefore, the partial pressure of HI (g) at equilibrium is approximately 6.46 atm.