Respuesta :
Answer:
Q = 0.50
No
Left
Explanation:
At a generic reversible equation
aA + bB ⇄ cC + dD
The reaction coefficient (Q) is the ratio of the substances concentrations:
[tex]Q = \frac{[C]^c*[D]^d}{[A]^a*[B]^b}[/tex]
Solids and liquid water are not considered in this calculus.
When the reaction achieves equilibrium (concentrations are constant), the Q value is named as Kc, which is the equilibrium constant of the reaction. If Q > Kc, it indicates that the concentration of the products is higher, so, the reaction must progress to the left and form more reactants; if Q < Kc, than the concentrations of the reactants, are higher, so, the reaction progress to the right.
In this case:
Q = [tex]\frac{[NO_2]^2}{[N_2O_4]}[/tex]
[tex]Q = \frac{0.50^2}{0.50}[/tex]
Q = 0.50
So, Q > Kc, the reaction is not at equilibrium and it progresses to the left.
The value of equilibrium constant for the reaction has been 0.5. The reaction has not been in equilibrium and process in the left direction.
The given balanced reaction has been:
[tex]\rm N_2O_4\;\leftrightharpoons\;2\;NO_2[/tex]
The value of the ratio of concentration of the product to the reactants at the equilibrium.
Computation of the Equilibrium Constant (Q)
The equilibrium constant for the given reaction has been:
[tex]Q=\rm \dfrac{[NO_2]^2}{[N_2O_5]}[/tex]
The concentration of the reactant and product at time t has been given as 0.5 M.
Substituting the values for the determination of Q:
[tex]Q=\dfrac{[0.5]^2}{[0.5]} \\Q=\dfrac{0.25}{0.5}\\Q=0.5[/tex]
The value of equilibrium constant for the reaction has been 0.5.
The value of Q greater than Kc has been the representation of the more product formation and the reaction in the left direction and vice versa.
The equal value of Q and Kc has been the representation of the equilibrium condition.
The value of Q calculated has been greater than given Kc value. Thus, the reaction has not been in equilibrium and process in the left direction.
Learn more about equilibrium constant, here:
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