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.