When 2.00 mol of SO₂Cl₂ is placed in a 2.00-L flask at 303 K, 56% of the SO₂Cl₂ decomposes to SO₂ and Cl₂: SO₂Cl₂(g) ⇌ SO₂(g) + Cl₂(g) (c) According to Le Châtelier's principle, would the percent of SO₂Cl₂ that decomposes increase, decrease or stay the same if the mixture were transferred to a 15.00-L vessel?

검증된 단계별 안내

Identify the initial conditions: 2.00 mol of SO_2Cl_2 in a 2.00-L flask, which means the initial concentration is 1.00 M.

Determine the change in concentration due to decomposition: 56% of SO_2Cl_2 decomposes, so calculate the concentration of SO_2 and Cl_2 formed.

Apply Le Châtelier's principle: Consider how a change in volume affects the equilibrium position. Increasing the volume decreases the pressure.

Analyze the reaction: The decomposition of SO_2Cl_2 results in an increase in the number of gas molecules (from 1 to 2), which means the reaction will shift to the right to counteract the decrease in pressure.

Conclude the effect: Since the reaction shifts to the right, the percent of SO_2Cl_2 that decomposes will increase when transferred to a 15.00-L vessel.

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주요 개념

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Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, resulting in constant concentrations of reactants and products. In this state, the system can respond to changes in conditions, such as concentration, temperature, or pressure, which can shift the equilibrium position according to Le Châtelier's principle.

Le Châtelier's Principle

Le Châtelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the system will adjust to counteract the change and restore a new equilibrium. This principle helps predict how the concentration of reactants and products will change in response to alterations in volume, pressure, or temperature.

Effect of Volume on Gas Equilibrium

When the volume of a gas reaction system is increased, the equilibrium will shift toward the side with more moles of gas to counteract the change. In the given reaction, the decomposition of SO2Cl2 produces two moles of gas (SO2 and Cl2) from one mole of SO2Cl2, indicating that increasing the volume will favor the formation of products.