2. Mixtures can be made of
1 answer:
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The equilibrium constant, Kc=0.026
<h3>Further explanation</h3>Given
1.72 moles of NOCI
1.16 moles of NOCI remained
2.50 L reaction chamber
Reaction
2NOCI(g) = 2NO(g) + Cl2(g).
Required
the equilibrium constant, Kc
Solution
ICE method
2NOCI(g) = 2NO(g) + Cl2(g).
I 1.72
C 0.56 0.56 0.28
E 1.16 0.56 0.28
Molarity at equilibrium :
NOCl :
NO :
Cl2 :
Answer:
0.011 mol/L
Explanation:
This can be solved with something called an ICE table.
I = initial
C = change
E = equilibrium
Initially, there is 0.04 M of N₂, 0.04 M of O₂, and 0 M of NO.
x amount of N₂ reacts. Since the stoichiometry is 1:1, x amount of O₂ also reacts. This produces 2x of NO.
After the reaction, there is 0.04-x of N₂, 0.04-x of O₂, and 2x of NO.
Here it is in table form:
Now we can use the equilibrium constant:
Kc = [NO]² / ( [N₂] [O₂] )
Substituting:
0.10 = (2x)² / ( (0.04 - x) (0.04 - x) )
Solving:
0.10 = (2x)² / (0.04 - x)²
√0.10 = 2x / (0.04 - x)
(√0.10) (0.04 - x) = 2x
(√0.10)(0.04) - (√0.10)x = 2x
(√0.10)(0.04) = 2x + (√0.10)x
(√0.10)(0.04) = (2 + √0.10)x
x = (√0.10)(0.04) / (2 + √0.10)
x = 0.0055
At equilibrium, the concentration of NO is 2x. So the answer is:
[NO] = 2x
[NO] = 0.011
The equilibrium concentration of NO is 0.011 mol/L.