Answer:
0.2024 M
Explanation:
For the decomposition reactio given, let's do an equilibrium chart. Let's call the initial concentration of NH₃ as C:
2NH₃(g) ⇄ N₂(g) + 3H₂(g)
C 0 0 Initial
-2x +x +3x Reacts (stoichiometry is 1:1:3)
C - 2x x 3x Equilibrium
3x = 0.252
x = 0.084 M
The equilibrium constant (Kc) is the multiplication of the concentrations of the products elevated by their coefficients, divided by the multiplication of reactants concentrations elevated by their coefficients.
Kc = ([H₂]³*[N₂])/([NH₃]²)
4.50 = [(0.252)³*(0.084)]/(C - 2*0.084)²
4.50 = 0.00533/(C - 0.168)²
4.50 = 0.00533/(C² - 0.336C + 0.028224)
4.50C² - 1.512C + 0.127008 = 0.00533
4.50C² - 1.512C + 0.121678 = 0
Solving the equation by a graphic calculator, for C > 0.168
C = 0.2024 M
The initial concentration of ammonia in the reaction flask is 0.2024 M.
The equilibrium constant is the ratio of the concentration of product to the reactant at equilibrium, raised to the stoichiometric coefficient.
From the balanced chemical equation, the equilibrium constant (Kc) for the reaction is expressed as:
[tex]Kc=\rm \dfrac{[N_2]\;[H_2]^3}{[NH_3]^2}[/tex]
Let the initial concentration of ammonia is C.
From the ICE table of the reaction attached, the equilibrium concentration of hydrogen is 0.252 M.
Thus,
[tex]3x=0.252\\x=0.084[/tex]
The equilibrium concentrations of reaction is given as:
Substituting the values for Kc:
[tex]\rm 4.50=\dfrac{[0.084]\;[0.252]^3}{[C-2(0.084)^3]}\\\\ 4.50=\dfrac{0.00533}{(C-0.168)^2}\\\\ C=0.2024[/tex]
The initial concentration of ammonia in the reaction flask is 0.2024 M.
Learn more about equilibrium constant, here:
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