Respuesta :
Answer:
Keq for this reaction is 0.87 (option a.)
Explanation:
The equilibrium is this:
N₂O₄  ⇄ 2N₂O₂
1 mol of Nâ‚‚Oâ‚„ decomposes to 2 moles Nâ‚‚Oâ‚‚
          N₂O₄  ⇄ 2N₂O₂
Initially    0.04m      -
React        x       2x
Initially I have 0.04 moles of Nâ‚‚Oâ‚„. Some amount (x) reacts to decompose and make 2 moles of Nâ‚‚Oâ‚‚; according to reaction, ratio is 1:2.
So if x amount of compound, reacted. I will have the double, as product decomposed.
          N₂O₄  ⇄ 2N₂O₂
Initially    0.04m      -
React        x       2x
Eq.        0.055m    2x
We have moles of Nâ‚‚Oâ‚„ in equilibrium, so we have to know how many amount has reacted.
(0.04 - 0.0055) = 0.0345
The double of this, is what we have in equilibrium of Nâ‚‚Oâ‚‚.
0.0345 .2 = 0.069
Let's make Kc:
[N₂O₂]² / [N₂O₄]
0.069² / 0.0055 = 0.87
Keq for this reaction is 0.87.
A. 0.87
Chemical reaction:
N₂O₄  ⇄ 2N₂O₂
1 mol of Nâ‚‚Oâ‚„ decomposes to 2 moles Nâ‚‚Oâ‚‚
         N₂O₄  ⇄  2N₂O₂
Initially    0.04m      -
React        x       2x
Initially I have 0.04 moles of Nâ‚‚Oâ‚„. Some amount (x) reacts to decompose and make 2 moles of Nâ‚‚Oâ‚‚; according to reaction, ratio is 1:2.
So if x amount of compound, reacted. I will have the double, as product decomposed.
          N₂O₄  ⇄  2N₂O₂
Initially    0.04m      -
React        x       2x
At Eq.      0.055m    2x
The moles of Nâ‚‚Oâ‚„ in equilibrium, so we have to know how many amount has reacted.
(0.04 - 0.0055) = 0.0345
The double of this, is what we have in equilibrium of Nâ‚‚Oâ‚‚.
0.0345 * 2 = 0.069
Calculation for kc:
[tex]k_c=\frac{[N_2O_2]^2}{[N_2O_4]} \\\\k_c=\frac{0.069^2}{0.0055}\\\\k_c=0.87[/tex]
Thus, the value of kc = 0.87. Hence, correct option is A.
Find more information about equilibrium constant here:
brainly.com/question/12270624
