Equilibrium Question

Question

For the unbalanced reaction

H2​(g)+NO(g)H2​O(g)+N2​(g)

the initial concentrations are 6.25 M H₂, 7.50 M NO, and no H₂O or N₂. At equilibrium, [N₂] = 1.50 M.

1.  Balance the equation using lowest-whole-number coefficients. Be sure to include states of matter in your equation.
2. Calculate the concentrations of H_2, NO, and H₂O at equilibrium.

     _____M H₂   

     _____M NO    

_____M H₂O   

3.Calculate the value of K under the reaction conditions at equilibrium.

Answer 1

2H₂(g) + 2NO(g) <==> 2H₂O(g) + N₂(g) .. balanced with whole number coefficients

To calculate concentrations @ equilibrium, we will set up an ICE table. And we are given [N₂]@ equilibrium

2H₂(g) + 2NO(g) <==> 2H₂O(g) + N₂(g)

6.25.........7.50.................0..............0.............Initial

-3.00......-3.00................+3.00......+1.50.......Change

3.25........4.50...................3.00........1.50.......Equilibrium

So concentrations at equilibrium are 3.25 M H₂, 4.50 M NO, 3.00 M H₂O and 1.50 M N₂

To calculate the value of K, solve the equilibrium expression as follows:

K = [H₂O]²[N₂] / [H₂]²[NO]²

K = (3.00)²(1.50) / (3.25)²(4.50)² = 13.5 / 0.522

K = 25.9