equilibrium pressure at 731 K for the reaction H2+I2-2HI initially the mixture contains 0.08592 each of H2 and I2 and 1.0atm.what is the pressure of HI at equilibrium?
Please rephrase your question. It isn't clear what you want AND you have no units on the 0.08592.
To find the equilibrium pressure of HI, you need to use the reaction equation and the concept of equilibrium constants.
The balanced chemical equation for the reaction is:
H2 + I2 ⇌ 2HI
The equilibrium constant expression (Kc) for this reaction can be written as:
Kc = [HI]^2 / [H2] * [I2]
Given that the initial moles of H2 and I2 are both 0.08592, and the initial total pressure is 1.0 atm, we can determine the initial concentrations of H2 and I2.
Concentration can be calculated by dividing the moles of a substance by the total volume.
Concentration (H2) = (0.08592 mol) / (total volume in liters)
Concentration (I2) = (0.08592 mol) / (total volume in liters)
Since only the total pressure is given and not the volume, we need to convert the initial total pressure to partial pressures of H2 and I2 using the ideal gas law.
Initially, the mixture contains equal moles of H2 and I2, so each gas will have a partial pressure of 0.5 atm (1.0 atm / 2).
Now, we can substitute these initial concentrations and partial pressures into the equilibrium constant expression:
Kc = [HI]^2 / (0.5)^2 (since [H2] and [I2] are equal)
Given that the equilibrium constant (Kc) for this reaction at 731 K is 49, you can rearrange the equation and solve for [HI]:
49 = [HI]^2 / (0.5)^2
Cross-multiply and solve for [HI]^2:
[HI]^2 = 49 * (0.5)^2
[HI]^2 = 49 * 0.25
[HI]^2 = 12.25
Taking the square root of both sides, we get:
[HI] = √12.25
[HI] ≈ 3.5
Therefore, the equilibrium pressure of HI at 731 K is approximately 3.5 atm.