The vapor pressure of water @ 80 degrees C is 0.467 atm. Find value of Kc for the reaction H2O(l) to H2O(g) @ this temp.

To find the value of Kc for the reaction H2O(l) ⇌ H2O(g) at 80 degrees Celsius, we first need to understand the relationship between Kc and the vapor pressure of water.

Kc, also known as the equilibrium constant, is a ratio of the concentrations of products to the concentrations of reactants at equilibrium. In the case of gases, Kc can also be expressed in terms of partial pressures. This is applicable to the reaction H2O(l) ⇌ H2O(g) because it involves the gas phase.

The vapor pressure of water is the pressure exerted by water vapor in equilibrium with liquid water at a particular temperature. In this case, the vapor pressure of water at 80 degrees Celsius is given as 0.467 atm.

Now, the equilibrium constant expression for the reaction is:

Kc = [H2O(g)] / [H2O(l)]

Since the reaction only involves water and the concentrations of pure liquids are constant, we can express Kc solely in terms of the partial pressure of water vapor:

Kc = [P(H2O(g))] / [P(H2O(l))]

Since the vapor pressure of water is a measure of the partial pressure of water vapor, we can substitute the given vapor pressure value into the equation:

Kc = [0.467 atm] / [P(H2O(l))]

However, because the reaction is at equilibrium, the partial pressure of water vapor is equal to the vapor pressure of water, which is given as 0.467 atm. Hence:

Kc = [0.467 atm] / [0.467 atm]

Simplifying this expression, we find:

Kc = 1

Therefore, the value of Kc for the reaction H2O(l) ⇌ H2O(g) at 80 degrees Celsius is 1.

To find the value of Kc for the reaction H2O(l) ⇌ H2O(g) at 80 degrees Celsius, we need to use the equilibrium constant expression.

The equilibrium constant expression for a reaction in terms of partial pressures is given by:

Kp = (P(H2O(g))) / (P(H2O(l)))

However, since we are given the vapor pressure of water (P(H2O(g))), we can directly use it to find Kc.

Kp is related to Kc by the following equation:

Kp = Kc * (RT)^Δn

Where:
R = Gas constant (0.0821 atm·L/mol·K)
T = Temperature in Kelvin (80 + 273.15 = 353.15 K)
Δn = Change in moles of gas (products - reactants)

In this case, there is no change in the number of moles of gas since both the reactant (liquid water) and the product (water vapor) have one mole of gas each. Hence, Δn = 0.

Therefore, the equation simplifies to:

Kp = Kc * (RT)^0
Kp = Kc

Now, to find Kc, we can equate Kp to the given value of the vapor pressure of water at 80 degrees Celsius:

0.467 atm = Kc

So, the value of Kc for the reaction H2O(l) ⇌ H2O(g) at 80 degrees Celsius is 0.467.

H2O(l) ==> H2O(g)

Kc = PH2O = 0.467.