The equilibrium constant for the chemical equation is K = 4.90 at 199 °C. Calculate the value of the Kc for the reaction at 199 °C.

N2(g)+3H2(g)=2NH3(g)

Kc=?????

I don't get the question.

Keq = Kc isn't it?

To calculate the value of Kc for the reaction at 199 °C, we need to understand the relationship between Kc and K.

The equilibrium constant Kc is related to the equilibrium constant K by the formula:

Kc = K(RT)^(Δn)

Where:
- Kc is the equilibrium constant in terms of concentrations
- K is the equilibrium constant in terms of activities
- R is the ideal gas constant (0.0821 L·atm/(mol·K))
- T is the temperature in Kelvin (199 + 273 = 472 K)
- Δn is the difference in the number of moles of gaseous products and gaseous reactants.

In the given chemical equation: N2(g) + 3H2(g) ⇌ 2NH3(g)
- The number of moles of gaseous products (NH3) is 2
- The number of moles of gaseous reactants (N2 + 3H2) is 1 + 3 = 4

Therefore, Δn = (2 - 4) = -2

Substituting the values into the formula, we get:

Kc = K(0.0821 L·atm/(mol·K))^(−2) = 4.90(0.0821 L·atm/(mol·K))^(−2)

Now we can calculate the value of Kc using this equation.