For the reaction, 2NO(g) ⇌ N2O2(g), Kc = 1.8 × 10–5 M–1 at 298 K. If [NO]0 = 0.50 M at 298 K, what is the concentration of N2O2(g) when the reaction reaches equilibrium?

Question

For the reaction, 2NO(g) ⇌ N2O2(g), Kc = 1.8 × 10–5 M–1 at 298 K.  If [NO]0 = 0.50 M at 298 K, what is the concentration of N2O2(g) when the reaction reaches equilibrium?

BlackTom AI · Accepted Answer

Let us analyze the equilibrium for the reaction: 2 NO ⇌ N2O2 with Kc = 1.8 × 10^–5 M^–1 at 298 K. The expression for Kc is Kc = [N2O2]/[NO]^2.

Option 1: 9.0 × 10^–6 M for [N2O2]. If [N2O2] = 9.0 × 10^–6 M, then let x = 9.0 × 10^–6. Since NO is consumed in a 2:1 ratio to form N2O2, [NO] at equilibrium would be approximately 0.50 − 2x ≈ 0.50 − 1.8 × 10^–5 ≈ 0.499982 M. The calculated Kc would be x / (0.499982)^2 ≈ 9.0 × 10^–6 / 0.249982 ≈ 3.6 × 10^–5, which is about twice the given Kc and clearly not equal to 1.8 × 10^–5. Thus this option is inconsistent with the equilibrium constant.

Option 2: 4.5 × 10^–6 M for [N2O2]. Set x = 4.5 × 10^–6. Then [NO] at equilibrium is 0.50 − 2x ≈ 0.50 − 9.0 × 10^–6 ≈ 0.499991 M. Compute Kc ≈ x / [NO]^2 ≈ (4.5 × 10^–6) / (0.499991)^2 ≈ (4.5 × 10^–6) / 0.249991 ≈ 1.8 × 10^–5 M^–1, which matches the given Kc. Since x is very small relative to 0.50 M, the approximation is valid and this value is consistent with the equilibrium expression.

Option 3: 0.060 M for [N2O2]. If [N2O2] = 0.060 M, then [NO] = 0.50 − 2(0.060) = 0.50 − 0.12 = 0.38 M. The resulting Kc would be 0.060 / (0.38)^2 ≈ 0.060 / 0.1444 ≈ 0.415, which is orders of magnitude larger than the given 1.8 × 10^–5. This clearly does not satisfy the equilibrium constant.

Option 4: 0.030 M for [N2O2]. If [N2O2] = 0.030 M, then [NO] = 0.50 − 2(0.030) = 0.50 − 0.060 = 0.44 M. Compute Kc ≈ 0.030 / (0.44)^2 ≈ 0.030 / 0.1936 ≈ 0.155, which again is far from 1.8 × 10^–5. This shows the pair (NO, N2O2) would not be at the specified equilibrium.

In summary, for this system, the value of [N2O2] that yields Kc = 1.8 × 10^–5 with the initial [NO]0 = 0.50 M is obtained by solving x / (0.50 − 2x)^2 = 1.8 × 10^–5, which, under the small-x approximation, gives x ≈ 1.8 × 10^–5 × (0.50)^2 = 1.8 × 10^–5 × 0.25 = 4.5 × 10^–6 M. This aligns with Option 2 and explains why the other options do not satisfy the equilibrium expression. The key idea is applying the ice table relationship and the equilibrium expression, then checking consistency for each candidate value against Kc.

Chem 1C General Chemis... Self Assessment Quiz 9

For the reaction, 2NO(g) ⇌ N2O2(g), Kc = 1.8 × 10–5 M–1 at 298 K. If [NO]0 = 0.50 M at 298 K, what is the concentration of N2O2(g) when the reaction reaches equilibrium?

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Chem 1C General Chemis... Self Assessment Quiz 9

For the reaction, 2NO(g) ⇌ N2O2(g), Kc = 1.8 × 10–5 M–1 at 298 K. If [NO]0 = 0.50 M at 298 K, what is the concentration of N2O2(g) when the reaction reaches equilibrium?

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类似问题

Methanol is manufactured by the following exothermic reaction. CO (g) + 2H2 (g) ⇌ CH3OH (g) Which pair of changes will favour the forward reaction?[Fill in the blank]

Chemical equilibrium is the state where chemical reactions are in balance and no longer occurring.

For the reaction N2(g) + 3 H2(g) 2 NH3(g) Kc = 0.0600 at a certain temperature. In an equilibrium mixture of the three gases, [NH3] = 0.242 M and [ H2] = 1.03 M. What is the concentration of N2 in this system?

Which of the following statements concerning equilibrium constants is true?

Terms relating to certain types of substances are conventionally omitted from equilibrium constant expressions. These substances are

At a certain temperature the equilibrium constant, Kc, equals 0.11 for the reaction: What is the equilibrium concentration of ICl if 0.75 mol of I2 and 0.75 mol of Cl2 are initially mixed in a 2.0-L flask?

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