Respuesta :
Explanation:
Since, the given reaction is as follows.
[tex]2NO(g) + Cl_{2} \rightleftharpoons 2NOCl[/tex]
Expression for [tex]Q_{c}[/tex] of this reaction is as follows.
[tex]Q_{c} = \frac{[NOCl]^{2}}{[NO]^{2}[Cl_{2}]}[/tex]
Putting the given values into the above formula as follows.
[tex]Q_{c} = \frac{[NOCl]^{2}}{[NO]^{2}[Cl_{2}]}[/tex]
= [tex]\frac{(9.50)^{2}}{(4.40 \times 10^{-2})^{2} \times (1.80 \times 10^{-3})}[/tex]
= [tex]\frac{90.25}{34.848 \times 10^{-7}}[/tex]
= [tex]2.59 \times 10^{-7}[/tex]
Thus, we can conclude that [tex]Q_{c}[/tex] for the experiment is [tex]2.59 \times 10^{-7}[/tex].
The equilibrium constant for a reaction is [tex]2.59*10^{-7}[/tex]. This can be found out by looking at reactants and products.
Balanced chemical reaction at equilibrium:
2NO(g) + Cl₂(g) ⇌ 2NOCl(g)
The equilibrium constant for this reaction can be represented as:
[tex]Q_c=\frac{[NOCl]^2}{[NO]^2[Cl_2]}[/tex]
Given:
The moles of chemical entities:
Moles of NO = 4.40 × 10−2 moles
Mole of Cl₂ = 1.80 × 10−3 moles
Moles of NaCl = 9.50 moles
On substituting the values:
[tex]Q_c=\frac{[NOCl]^2}{[NO]^2[Cl_2]}\\\\Q_c=\frac{9.50^2}{(4.40*10^{-2})^2(1.80*10^{-3})^2} \\\\Q_c=2.59*10^{-7}[/tex]
Thus, we can conclude that for the experiment, the value for equilibrium constant is [tex]2.59*10^{-7}[/tex].
Find more information about equilibrium constant here:
brainly.com/question/12858312
