At 860.0 deg celsius, the equilibrium constant Kp for the synthesis of ammonia (below) is 3.35 * 10^-5. What is the value of Kc?

N2(g) + 3H2(g) <--> 2NH3(g)

Kp = Kc(RT)^delta n

where delta n = nproducts-nreacrtants.

To find the value of Kc, we need to use the relationship between Kp and Kc for a gas-phase reaction. The equilibrium constant Kp is related to Kc through the equation:

Kp = Kc(RT)^(Δn)

Where:
- Kp is the equilibrium constant at constant pressure,
- Kc is the equilibrium constant at constant concentration,
- R is the gas constant (0.0821 L·atm/(mol·K)),
- T is the temperature in Kelvin, and
- Δn is the difference in the number of moles of gaseous products and reactants.

In the given reaction, the Δn can be calculated as follows:
- For reactants: 1 mole of N2 and 3 moles of H2, giving a total of 4 moles of reactants.
- For products: 2 moles of NH3, giving a total of 2 moles of products.

Hence, Δn = (2 - 4) = -2.

Now, let's plug in the known values into the equation to find Kc:

Kp = 3.35 * 10^-5
R = 0.0821 L·atm/(mol·K)
T = temperature in Kelvin

To convert 860.0 degrees Celsius to Kelvin, we use the equation: T(K) = T(°C) + 273.15.

T(K) = 860 + 273.15 = 1133.15 K

Now, we can substitute the values into the equation and solve for Kc:

3.35 * 10^-5 = Kc*(0.0821 L·atm/(mol·K))*(1133.15 K)^(-2)

Simplifying the equation:

3.35 * 10^-5 = Kc*(0.0821 L·atm/(mol·K))*(1/(1133.15 K)^2)

Rearranging and solving for Kc:

Kc = (3.35 * 10^-5)/(0.0821 * (1133.15)^2)

Using a calculator, we find:

Kc ≈ 3.91 × 10^-2

Therefore, the value of Kc for the synthesis of ammonia at 860.0 degrees Celsius is approximately 3.91 × 10^-2.