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

Step-by-step solution

For the given reversible reaction: N _ 2(g) + 3 H _ 2(g) 2 NH _ 3(g) The change in the number of moles of gaseous species is: n_g = moles of gaseous products - moles of gaseous reactants n_g = 2 - (1 + 3) = -2 The relationship between K_p and K_c is given by: K_p = K_c(RT)^ n_g Given values are: K_p = 1.44 10⁻⁵ R = 0.082 L atm K ⁻¹ mol ⁻¹ T = 500^ C = 500 + 273 = 773 K Substituting these values into the equation: 1.44 10⁻⁵ = K_c(0.082 773)⁻² Rearranging to solve for K_c : K_c = 1.44 10⁻⁵ (0.082 773)⁻² Answer: 1.44