Answer:
Kc for this equilibrium is 2.30*10⁻⁶
Explanation:
Equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction and the concentrations of reactants and products are held constant.
Being:
aA + bB ⇔ cC + dD
the equilibrium constant Kc is defined as:
![Kc=\frac{[C]^{c}*[D]^{d} }{[A]^{a} *[B]^{b} }](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BC%5D%5E%7Bc%7D%2A%5BD%5D%5E%7Bd%7D%20%20%7D%7B%5BA%5D%5E%7Ba%7D%20%2A%5BB%5D%5E%7Bb%7D%20%7D)
In other words, the constant Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients. Kc is constant for a given temperature, that is to say that as the reaction temperature varies, its value varies.
In this case, being:
2 NH₃(g) ⇔ N₂(g) + 3 H₂(g)
the equilibrium constant Kc is:
![Kc=\frac{[N_{2} ]*[H_{2} ]^{3} }{[NH_{3} ]^{2} }](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BN_%7B2%7D%20%5D%2A%5BH_%7B2%7D%20%5D%5E%7B3%7D%20%20%7D%7B%5BNH_%7B3%7D%20%5D%5E%7B2%7D%20%7D)
Being:
- [N₂]= 0.0551 M
- [H₂]= 0.0183 M
- [NH₃]= 0.383 M
and replacing:

you get:
Kc= 2.30*10⁻⁶
<u><em>Kc for this equilibrium is 2.30*10⁻⁶</em></u>
<span>We have ground strate configurations of electrons,if electrons are filled in order of increasing energy. When there are electrons are in higher orbitals, we have an atom in an excited state.
B, and C are excited states.
In B, 2 electrons can fit in the 4s orbital, and that should fill fully before the 4p orbitals.
In C, the same is true for 5s and 5p
In D, this is not an excited state because 4s fills before 3d</span>