2SO2 (g) + O2 (g) ⇌ 2SO3 (g) For this exothermic reaction, the equilibrium constant, K = 781. If, after equilibrium has been reached, the volume of the vessel is suddenly increased the result will be:

Question

2SO2 (g) + O2 (g) ⇌ 2SO3 (g) For this exothermic reaction, the equilibrium constant, K = 781. If, after equilibrium has been reached, the volume of the vessel is suddenly increased the result will be:

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The reaction given is 2SO2(g) + O2(g) ⇌ 2SO3(g), and it is exothermic with K = 781 at the temperature of interest. When the volume of the reaction vessel is increased at constant temperature, the total pressure decreases and the system experiences a stress that may be relieved by shifting the equilibrium toward the side with more gas molecules or fewer gas molecules depending on the change in moles. In this reaction, the left side (reactants) has 3 moles of gas (2 from SO2 and 1 from O2) while the right side (products) has 2 moles of gas (2 from SO3). Since increasing volume favors the side with more moles of gas to counteract the drop in pressure, this situation actually works against the general “more moles means shift to left” intuition unless temperature or other conditions specify otherwise; however, the key is to compare mole counts: the right side has fewer moles of gas than the left. Therefore, the system shifts toward the side with fewer moles of gas to minimize the effect of the volume increase, which is the forward direction producing more SO3. Option analysis:
- a. unpredictable: Not correct, because the effect of a volume increase on this gas-phase equilibrium is predictable using Le Châtelier’s principle and the difference in gas moles.
- b. reaction in the forward direction: This aligns with shifting toward fewer moles of gas when volume increases, making this the correct directional change under the given conditions.
- c. reaction in the backward direction: This would be chosen if the left side had fewer moles of gas than the right, or if the system favored more gas molecules upon expansion; here it does not.
- d. no change in the forward or backward reactions: Not correct, since a volume change at constant temperature imposes a stress that alters the equilibrium position.

2SO2 (g) + O2 (g) ⇌ 2SO3 (g) For this exothermic reaction, the equilibrium constant, K = 781. If, after equilibrium has been reached, the volume of the vessel is suddenly increased the result will be:

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2SO2 (g) + O2 (g) ⇌ 2SO3 (g) For this exothermic reaction, the equilibrium constant, K = 781. If, after equilibrium has been reached, the volume of the vessel is suddenly increased the result will be:

查看解析

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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?

Consider the equilibrium system C(s) + CO2(g) ⇔2CO(g) If C is removed, the equilibrium will _____, and if CO is added, the equilibrium will _____.

Consider the equilibrium reaction: CO (g) + H2 (g) ⇌ C (s) + H2O (g), ΔHr = –131.3 kJ mol–1 Making use of Le Chatelier’s principle, identify which of the following will happen if the temperature of the system is increased.

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