For the reversible reaction,
\(\text{N}_{2(g)} + 3\text{H}_{2(g)} \rightleftharpoons 2\text{NH}_{3(g)}\)
When the partial pressure is measured in atmosphere, the value of \(K_p\) at \(500^\circ\text{C}\) is \(1.44 \times 10^{-5}\). The value of \(K_c\) when the concentration is expressed in mol L\(^{-1}\) is:

For the given reversible reaction:
\(\text{N}_{2(g)} + 3\text{H}_{2(g)} \rightleftharpoons 2\text{NH}_{3(g)}\)

The change in the number of moles of gaseous species is:
\(\Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants}\)
\(\Delta n_g = 2 - (1 + 3) = -2\)

The relationship between \(K_p\) and \(K_c\) is given by:
\(K_p = K_c(RT)^{\Delta n_g}\)

Given values are:
\(K_p = 1.44 \times 10^{-5}\)
\(R = 0.082 \text{ L atm K}^{-1} \text{ mol}^{-1}\)
\(T = 500^\circ\text{C} = 500 + 273 = 773 \text{ K}\)

Substituting these values into the equation:
\(1.44 \times 10^{-5} = K_c(0.082 \times 773)^{-2}\)

Rearranging to solve for \(K_c\):
\(K_c = \dfrac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}}\)

Answer: \(\dfrac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}}\)