For the unbalanced reaction H2(g)+NO(g) ---> H2O(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.
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.
For the unbalanced reaction H2(g)+NO(g) ---> H2O(g)+N2(g): the initial concentrations are 5.25 M H2, 9.00 M...
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