# Exercise 3 | Chemical Equilibrium

A reaction has a two steps mechanism with the following equilibrium constant:

K1 $\frac{\left[{\mathrm{HO}}^{-}\right]\left[{{\mathrm{HCO}}_{3}}^{-}\right]}{\left[{{\mathrm{CO}}_{3}}^{2-}\right]}$  and K2$\frac{\left[{{\mathrm{HCO}}_{3}}^{-}\right]}{\left[{\mathrm{HO}}^{-}\right]\left[{\mathrm{H}}_{2}{\mathrm{CO}}_{3}\right]}$

Write the equilibrium constant expression and the chemical equation of the overall reaction.

Koverall = K1 x K2

Koverall$\frac{\left[{\mathrm{HO}}^{-}\right]\left[{{\mathrm{HCO}}_{3}}^{-}\right]}{\left[{{\mathrm{CO}}_{3}}^{2-}\right]}$ x $\frac{\left[{{\mathrm{HCO}}_{3}}^{-}\right]}{\left[{\mathrm{HO}}^{-}\right]\left[{\mathrm{H}}_{2}{\mathrm{CO}}_{3}\right]}$

Koverall = $\frac{{\left[{{\mathrm{HCO}}_{3}}^{-}\right]}^{2}}{\left[{{\mathrm{CO}}_{3}}^{2-}\right]\left[{\mathrm{H}}_{2}{\mathrm{CO}}_{3}\right]}$

2 HCO3(g) $⇄$ CO32- (g) + H2CO(g)