# Question #05cf4

Apr 29, 2015

The equilibrium constant will be very large because that reaction will be very, very close to going to completion, which means that you'll have significantly more product than reactants at equilibrium.

$2 {H}_{2 \left(g\right)} + {O}_{2 \left(g\right)} r i g h t \le f t h a r p \infty n s 2 {H}_{2} {O}_{\left(g\right)}$

By definition, the equilibrium constant will be

$K = \frac{{\left[{H}_{2} O\right]}^{2}}{{\left[{H}_{2}\right]}^{2} \cdot \left[{O}_{2}\right]}$

Essentially, the hydrogen will react completely with oxygen. By comparison, the amounts of reactants left will be insignificant compared with the amount of water formed.

If you get much more product than reactants at equilibrium, $K$ will be very large because you divide the square of a large(r) number by the product of smaller numbers.

$\textcolor{g r e e n}{K > {10}^{3}}$