At 700 K, the reaction below has an Kp value of 54. An equilibrium mixture at this temperature was found to contain 0.933 atm of H2 and 2.1 atm of HI. Calculate the equibrium pressure of I2. H2(g) + I2(g) <=> 2 HI(g). Enter to 2 decimal place.

See explanation below

Explanation:

In this case, we have the equilibrium reaction which is:

H₂ + I₂ <> 2HI       Kp = 54

Now, we have the partial pressures of each element in equilibrium, therefore, we can use the expression of equilibrium in this case to calculate the remaining pressure:

Kp = PpHI² / PpH₂ * PpI₂

Solving for the partial pressure of iodine:

PpI₂ = PpHI² / PpH₂ * Kp

Replacing the given values, we have:

PpI₂ = (2.1)² / 0.933 * 54

PpI₂ = 4.41 / 50.382

PpI₂ = 0.088 atm

hi my dude or dudette

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.