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.
pBr2 = ?
n = 0.960
R = 0.08206
T = 273 + 1000 = 1273K
p = nRT/V = (0.960)*0.08206 x 1273/1 = 100.28 BUT I said that was approx and you should go through it with more accuracy. I would round that to 100.
pH2 = (1.28*0.08206 x 1273/1 = 128.49 with the same recommendation. I would round that to 128. I suspect you may have used a slightly different value for R which might account for the slight difference in pBr2 but for pH2 I suspect you just punched in the wrong numbers. OR you may have used a different value for R. Some round the 0.08206 (I think it's 0.082057) off to 0.0821
Hello, this is a few years late, but you should not be adding another 273 to 1000K that is already in kelvin. You should only add 273 when going from Celcius to Kelvin. This in effect alters the rest of your numbers.
To calculate the partial pressures at equilibrium, we need to use the 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.
First, let's calculate the initial moles of H2 gas:
Given that we have 1.23 mol of H2 in a 1.00 L reaction vessel, the initial concentration of H2 is 1.23 mol/L.
Next, let's calculate the initial pressure of H2:
Using the ideal gas equation, PV = nRT, we can rearrange it to solve for P:
P = (n/V) * RT
Plugging in the values, we get:
P = (1.23 mol / 1.00 L) * R * 1000 K
Using the value of the ideal gas constant, R = 0.0821 L·atm/(mol·K), we can calculate the initial pressure of H2:
P = (1.23 mol / 1.00 L) * 0.0821 L·atm/(mol·K) * 1000 K
P ≈ 100.26 atm
So, the initial partial pressure of H2 is approximately 100.26 atm.
Now, let's calculate the initial moles of Br2:
Given that we have 0.960 mol of Br2, the initial concentration of Br2 is 0.960 mol/L.
Next, let's calculate the initial pressure of Br2:
Using the ideal gas equation, PV = nRT, we can rearrange it to solve for P:
P = (n/V) * RT
Plugging in the values, we get:
P = (0.960 mol / 1.00 L) * R * 1000 K
Using the value of the ideal gas constant, R = 0.0821 L·atm/(mol·K), we can calculate the initial pressure of Br2:
P = (0.960 mol / 1.00 L) * 0.0821 L·atm/(mol·K) * 1000 K
P ≈ 78.96 atm
So, the initial partial pressure of Br2 is approximately 78.96 atm.
To find the partial pressure of HBr at equilibrium, we need to use the equilibrium constant Kp and the stoichiometric coefficients of the balanced chemical equation.
Given that Kp = 2.1×10^6, we can use the equation:
Kp = (P_HBr^2) / (P_H2 * P_Br2)
Plugging in the values we know:
2.1×10^6 = (P_HBr^2) / (100.26 atm * 78.96 atm)
Simplifying the equation, we find:
(P_HBr^2) = 2.1×10^6 * (100.26 atm * 78.96 atm)
Taking the square root of both sides, we get:
P_HBr = sqrt(2.1×10^6 * (100.26 atm * 78.96 atm))
Calculating the value, we find:
P_HBr ≈ 153730.29 atm
So, the partial pressure of HBr at equilibrium is approximately 153730.29 atm.
Thus, the partial pressures at equilibrium are as follows:
PH2 ≈ 100.26 atm
PBr2 ≈ 78.96 atm
PHBr ≈ 153730.29 atm
To determine the partial pressures of H2, Br2, and HBr at equilibrium, we need to use the given information and the equilibrium expression for the reaction: H2(g) + Br2(g) ⇌ 2HBr(g)
1. Start by determining the initial pressures of H2 and Br2:
Remember that in this case, we can assume the initial pressure of H2 is equal to the partial pressure of H2 at equilibrium (pH2).
Let's denote the initial pressure of Br2 as pBr2.
2. Use the ideal gas law, PV = nRT, to solve for the initial pressure of H2 (pH2):
PV = nRT
pH2 * V = nH2 * R * T
pH2 = (nH2 * R * T) / V
Given:
nH2 = 1.23 mol (moles of H2)
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
T = 1000 K (temperature in Kelvin)
V = 1.00 L (volume)
pH2 = (1.23 mol * 0.0821 L·atm/(mol·K) * 1000 K) / 1.00 L = 101.19 atm
So the initial pressure of H2 (pH2) is approximately 101.19 atm.
3. Similarly, use the ideal gas law to solve for the initial pressure of Br2 (pBr2):
Given:
nBr2 = 0.960 mol (moles of Br2)
pBr2 = (0.960 mol * 0.0821 L·atm/(mol·K) * 1000 K) / 1.00 L = 98.97 atm
So the initial pressure of Br2 (pBr2) is approximately 98.97 atm.
4. Now, use the stoichiometry of the reaction to determine the change in pressure for each species at equilibrium:
From the balanced equation: 1 mol of H2 produces 2 mol of HBr.
At equilibrium, let's assume the change in pressure for HBr is p, which means the pressure of HBr is 2p.
The change in pressure for Br2 and H2 will be -p (decreasing) due to their stoichiometric relationship.
5. Finally, use the equilibrium expression and the given equilibrium constant to set up the equation:
Kp = (pHBr^2) / (pH2 * pBr2)
Given:
Kp = 2.1 × 10^6
pH2 = 101.19 atm (from step 2)
pBr2 = 98.97 atm (from step 3)
pHBr = 2p (from step 4)
Substitute the values into the equation and solve for p:
2.1 × 10^6 = (2p)^2 / (101.19 * 98.97)
2.1 × 10^6 = 4p^2 / (101.19 * 98.97)
Solve for p using algebraic manipulation.
After solving this equation more accurately, Dr. Bob obtained approximately 100 for pBr2 and 128 for pH2 initially. You mentioned you got different values, so please check your calculations again to ensure accuracy.