This is an incomplete question, here is the complete question.
[tex]CO(g)+NH_3(g)\rightleftharpoons HCONH_2(g)[/tex], Kc = 0.810
If a reaction vessel initially contains only CO and NH₃ at concentrations of 1.00 M and 2.00 M, respectively, what will the concentration of HCONH₂ be at equilibrium?
Answer : The concentration of HCONH₂ at equilibrium is, 3.69 M
Solution : Given,
Initial concentration of [tex]CO[/tex] and [tex]NH_3[/tex] is, 1.00 M and 2.00 M
The given equilibrium reaction is,
[tex]CO+NH_3\rightleftharpoons HCONH_2[/tex]
Initially 1.00 2.00 0
At equilibrium (1.00-x) (2.00-x) x
The expression of [tex]K_c[/tex] will be,
[tex]K_c=\frac{[HCONH_2]}{[CO][NH_3]}[/tex]
[tex]0.810=\frac{(x)}{(1.00-x)\times (2.00-x)}[/tex]
By solving the terms, we get:
x = 3.69 M
Thus, the concentration of HCONH₂ at equilibrium is, 3.69 M
