Question

For the unbalanced reaction H2​(g)+NO(g) ---> H2​O(g)+N2​(g):

the initial concentrations are 5.25 M H2, 9.00 M NO, and no H2O or N2. At equilibrium, [N2] = 1.75 M.

A) Balance the equation using lowest-whole-number coefficients. Be sure to include states of matter in your equation.

B) Calculate the value of K under the reaction conditions at equilibrium.

Answer 1

H2(g) + NO(g) ----------- H2O(g) + N2(g)

The balanced equation i s

               2 H2(g) + 2NO(g) ----------- 2H2O(g) + N2(g)

Initial concentration of H2= 5.25M

                                  [NO] = 9.00M

at equilibrium [N2] = 1.75M

a)

The balanced equation is

                                      2 H2(g) + 2NO(g) ----------- 2H2O(g) + N2(g)

B)

                                    2 H2(g) + 2NO(g) ----------- 2H2O(g) + N2(g)

Initial                               5.25         9.00                    0               0

chnage                            -2x           -2x                     +2x             +x

equilibrium                  5.25-2x        9.00-2x               +2x               +x ( 1.75)

according to given data

at equilibrium [N2] = x= 1.75M

x= 1.75M

at equilibrium

[H2] = 5.25 - 2x = 5.25 - 2(1.75) = 1.75M

[NO] = 9.00-2x= 9.00 - 2(1.75) = 5.5M

[H2O] = 2x = 2 x1.75 = 3.5M

K= [H2O]^2 [N2]/ [H2[^2[NO]^2

K = ( 3.5)^2 x 1.75 / (1.75)^2 x(5.5)^2

K= 21.4375/92.641

K = 0.231

Equilibrium constant = K = 0.231.