The initial pressures for I2 (g), H2(g), and HI(g) were Pi2 = 0.100 atm, Ph2 = 0.200 atm, and Phi = 0 atm, respectively. After the system came to equilibrium, the pressure of I2 (g) became very low, PI2 = 1.00 x 10^-5 atm. Calculate the equilibrium constant Kp for this reaction
for WHAT reaction?
The reaction is I2(g) + H2(g) -> 2HI(g)
...........H2 + I2 ==> 2HI
initial....0.2..0.1......0
change.....-x....-x.....+2x
equil...........1E-5........
I would see it this way. We reacted ALMOST 0.1 atm of I2 to leave such a small value at equilibrium. That means we used up 0.1 of the H2 to leave 0.1 at equil and we formed 0.2 atm HF. Substitute and solve for Kp. You want to substitute 1E-5 for I2.
So instead of using 0.1 for I2, I want to use 1E-5 for I2 when solving for KP?
Or do I solve for x, using the given pressure of I2 at equilibrium,
so should it be 1E-5 = 0.1 - x? (solve for x)
If you wish to solve for x then x =0.099999 but since that is so close to 0.1, you will call it 0.1 as the p of I2 used.
Kp = pHI^2/pH2*pI2
pI2 in this set up is 1E-5.
Hmm I'm still confused at what to plug in for pHI^2 and pH2 for solving Kp
Nevermind I got it, with x = 0.099999 then I just use what I have in the ICE chart and plug it in to solve for Kp,
Thanks for all the help!
To calculate the equilibrium constant (Kp) for this reaction, you can use the ideal gas equation and the expression for Kp.
The balanced equation for the reaction is:
I2 (g) + H2(g) ⇌ 2HI(g)
First, let's determine the equilibrium partial pressures for I2, H2, and HI.
Given:
Pi2 = 0.100 atm (initial partial pressure of I2)
Ph2 = 0.200 atm (initial partial pressure of H2)
Phi = 0 atm (initial partial pressure of HI)
PI2 = 1.00 x 10^-5 atm (equilibrium partial pressure of I2)
Since the stoichiometric coefficient for I2 is 1, the equilibrium partial pressure of I2 is PI2 = 1.00 x 10^-5 atm.
To calculate the equilibrium partial pressure for H2, we need to use the stoichiometry of the balanced equation.
Using the balanced equation:
I2 (g) + H2(g) ⇌ 2HI(g)
The initial pressure of I2 is used up to form HI, so we subtract the change in pressure from the initial pressure of I2:
Change in pressure of I2 = Pi2 - PI2 = 0.100 atm - 1.00 x 10^-5 atm = 0.09999 atm
Since one mole of I2 produces two moles of HI, the change in pressure of H2 is half of the change in pressure of I2:
Change in pressure of H2 = (1/2) * (Change in pressure of I2) = (1/2) * 0.09999 atm = 0.049995 atm
The equilibrium partial pressure of H2 is then given by:
Ph2 (equilibrium) = Ph2 (initial) - Change in pressure of H2 = 0.200 atm - 0.049995 atm = 0.150005 atm
As for HI, there is no initial pressure (Phi = 0 atm), so the equilibrium partial pressure of HI is simply equal to the change in pressure of I2:
Phi (equilibrium) = Change in pressure of I2 = 1.00 x 10^-5 atm
Now, we can write the expression for Kp using the equilibrium partial pressures:
Kp = (Phi (equilibrium))^2 / (Pi2 - PI2) * (Ph2 (equilibrium))^2
Substituting the values:
Kp = (1.00 x 10^-5 atm)^2 / (0.100 atm - 1.00 x 10^-5 atm) * (0.150005 atm)^2
Calculating the value:
Kp = 1.00 x 10^-10 atm^2 / (9.999 x 10^-2 atm) * 0.022500300025 atm^2
Simplifying:
Kp = 1.00 x 10^-10 / 9.999 x 10^-2 * 0.022500300025 atm^2
Kp = 2.25 x 10^-7 atm
Therefore, the equilibrium constant Kp for this reaction is 2.25 x 10^-7 atm.