A 0.960mol quantity of Br2 is added to a 1.00 L reaction vessel that contains 1.23mol of H2 gas at 1000 K. What are the partial pressures of H2, Br2, and HBr at equilibrium?
At 1000 K, Kp=2.1×106 and ΔH∘ = -101.7 kJ for the reaction H2(g)+Br2(g)⇌2HBr(g).
Use PV = nRT to solve for pH2 initial.
Use PV = nRT to solve for p
I obtained approx 100 for pBr2 and 128 for pH2 initially but you need to go through and do it more accurately.
...........H2 + B2 ==> 2HBr
I..........128..100.....0
C...........-p..-p......+2p
E.......128-p..100-p....+2p
Substitute the E line into the Kp expression and solve for p followed by working out partial pressures of each. Post your work if you run into trouble.
How did you get 100 and 128? I got 101 and 78.
A 0.960mol quantity of Br2 is added to a 1.00 L reaction vessel that contains 1.23mol of H2 gas at 1000 K. What are the partial pressures of H2, Br2, and HBr at equilibrium?
At 1000 K, Kp=2.1×106 and ΔH∘ = -101.7 kJ for the reaction H2(g)+Br2(g)⇌2HBr(g).
Dr Bob, you said:
Use PV = nRT to solve for pH2 initial.
Use PV = nRT to solve for p
I obtained approx 100 for pBr2 and 128 for pH2 initially but you need to go through and do it more accurately.
...........H2 + B2 ==> 2HBr
I..........128..100.....0
C...........-p..-p......+2p
E.......128-p..100-p....+2p
How did you get 100 and 128? I got 101 and 78.
To determine the partial pressures of H2, Br2, and HBr at equilibrium, you will need to use the information given about the reaction and its equilibrium constant Kp.
First, let's set up the balanced equation for the reaction:
H2(g) + Br2(g) ⇌ 2HBr(g)
According to the stoichiometry of the balanced equation, the molar ratio between H2 and HBr is 1:2. This means that for every 1 mole of H2 that reacts, 2 moles of HBr are formed.
Now, let's calculate the initial partial pressures of H2, Br2, and HBr using the provided information.
- For H2:
Given that the initial number of moles of H2 in the reaction vessel is 1.23 mol, and the volume is 1.00 L, we can use the ideal gas law to determine the initial partial pressure of H2. The ideal gas law equation is:
PV = nRT
Where:
P = partial pressure (unknown)
V = volume (1.00 L)
n = number of moles (1.23 mol)
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (1000 K)
Rearranging the equation to solve for P, we have:
P = (nRT) / V = (1.23 mol * 0.0821 L·atm/mol·K * 1000 K) / 1.00 L = 100.776 atm
So the initial partial pressure of H2 is 100.776 atm.
- For Br2:
Given that the initial number of moles of Br2 added to the reaction vessel is 0.960 mol, and the volume is 1.00 L, we can again use the ideal gas law to determine the initial partial pressure of Br2:
P = (nRT) / V = (0.960 mol * 0.0821 L·atm/mol·K * 1000 K) / 1.00 L = 79.9392 atm
So the initial partial pressure of Br2 is 79.9392 atm.
Now, to determine the equilibrium partial pressures of H2, Br2, and HBr, we can use the equilibrium constant expression:
Kp = (P_HBr)^2 / (P_H2 * P_Br2)
Given that Kp = 2.1×10^6, we can rearrange the equation to solve for the equilibrium partial pressures:
(P_HBr)^2 = Kp * P_H2 * P_Br2
Taking the square root of both sides:
P_HBr = √(Kp * P_H2 * P_Br2)
Plugging in the values we calculated earlier:
P_HBr = √(2.1×10^6 * 100.776 atm * 79.9392 atm) = 28030.09 atm
Therefore, the equilibrium partial pressure of HBr is 28030.09 atm.
Since the reaction is a stoichiometric relationship of 1:1 between Br2 and HBr, the partial pressure of Br2 at equilibrium will be equal to the partial pressure of HBr:
P_Br2 = P_HBr = 28030.09 atm
In summary, the partial pressures at equilibrium are as follows:
- Partial pressure of H2: 100.776 atm
- Partial pressure of Br2: 28030.09 atm
- Partial pressure of HBr: 28030.09 atm