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Consider this equilibrium system: $$ \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) $$ Predict how the equilibrium position would change if (a) \(\mathrm{Cl}_{2}\) gas were added to the system, (b) \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) were removed from the system, (c) \(\mathrm{SO}_{2}\) were removed from the system. The temperature remains constant

Short Answer

Expert verified
(a) The equilibrium will shift to the right, resulting in the production of more \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\). \n(b) The equilibrium will shift to the right to counteract the removal of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\). \n(c) The equilibrium will shift to the left to increase the production of \(\mathrm{SO}_{2}\).

Step by step solution

01

Adding Cl2 to the system

According to Le Chatelier's Principle, if \(\mathrm{Cl}_{2}\) is added to system, the equilibrium will shift to the right to counteract this change. This is because the reaction will try to remove the excess \(\mathrm{Cl}_{2}\) by using it to form more \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\). Therefore, adding more \(\mathrm{Cl}_{2}\) would cause the reaction to generate more \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) to reach equilibrium again.
02

Removing SO2Cl2 from the system

If \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is removed from the system, the reaction would shift towards the right to increase the concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\). The shift to the right will increase the concentrations of the product and decrease the concentrations of reactants effectively nullifying the effect of removing \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) to reach equilibrium.
03

Removing SO2 from the system

On the contrary, if \(\mathrm{SO}_{2}\) is removed from the system, the equilibrium will shift to the left to increase the concentration of \(\mathrm{SO}_{2}\). This means the reaction will favor the backward reaction and the \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) will break down into \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Le Chatelier's Principle
Le Chatelier's Principle is a foundational concept in chemistry which predicts how a change in conditions can affect the equilibrium of a chemical system. When a system at equilibrium is disturbed, it will adjust in a way that counteracts the disturbance and re-establishes equilibrium. Think of it like a seesaw that adjusts itself to stay balanced, despite weights being added or removed on either side.

For example, if the concentration of reactants is increased, the system will shift to produce more products to reduce this sudden excess. Similarly, if a product is removed from the system, the equilibrium will shift to produce more of that product in response. Adjustments can also occur in response to changes in temperature or pressure. These adjustments are the system's way of following Le Chatelier's Principle to re-balance itself.
Equilibrium Shift
An equilibrium shift occurs when a reversible chemical reaction responds to a change in conditions, moving the balance between reactants and products. It's not about completely changing a reaction's direction but about adjusting the amount of reactants and products until they reach a new state of balance.

Imagine a pair of scales with weights on each side. If we suddenly add more weight to one side, the scales will tip. In a chemical reaction, the 'adding of weight' could be increasing the concentration of a reactant or removing a product. The reaction 'tips' by favoring the formation of products or reactants until a new equilibrium is established. In response to various changes, such as concentration alterations as illustrated in the exercise, the shifts help the system find a new 'balance point' while still honoring the principle of equilibrium.
Reactants and Products
Reactants and products are the starting and resulting substances of a chemical reaction, respectively. At the start of a reaction, you have the reactants which interact to form the products. However, in a dynamic equilibrium, this process is reversible. The products can recombine to form reactants again.

In a chemical equilibrium, the rates of the forward and reverse reactions are equal, leading to a constant ratio of reactants to products, even though individual molecules are continuously reacting. The equilibrium does not mean the reactants and products are in equal amounts, but that the rate at which reactants turn into products is equal to the rate at which products revert into reactants. When changes occur in the system, the proportion of reactants and products shifts until a new state of balance is found, as per Le Chatelier's Principle.

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Most popular questions from this chapter

At \(25^{\circ} \mathrm{C}\), the equilibrium partial pressures of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) are \(0.15 \mathrm{~atm}\) and \(0.20 \mathrm{~atm}\), respectively. If the volume is doubled at constant temperature, calculate the partial pressures of the gases when a new equilibrium is established.

The "boat" form and "chair" form of cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{12}\right)\) interconverts as shown here: In this representation, the \(\mathrm{H}\) atoms are omitted and a \(\mathrm{C}\) atom is assumed to be at each intersection of two lines (bonds). The conversion is first order in each direction. The activation energy for the chair \(\longrightarrow\) boat conversion is \(41 \mathrm{~kJ} / \mathrm{mol}\). If the frequency factor is \(1.0 \times 10^{12} \mathrm{~s}^{-1}\), what is \(k_{1}\) at \(298 \mathrm{~K}\) ? The equilibrium constant \(K_{c}\) for the reaction is \(9.83 \times 10^{3}\) at \(298 \mathrm{~K}\).

Write the equilibrium constant expressions for \(K_{\mathrm{c}}\) and \(K_{P}\), if applicable, for these reactions: (a) \(2 \mathrm{NO}_{2}(g)+7 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\) (b) \(2 \mathrm{ZnS}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{ZnO}(s)+2 \mathrm{SO}_{2}(g)\) (c) \(\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}(a q) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}(a q)+\mathrm{H}^{+}(a q)\)

Consider the reaction $$ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ At \(430^{\circ} \mathrm{C}\), an equilibrium mixture consists of 0.020 mole of \(\mathrm{O}_{2}, 0.040\) mole of \(\mathrm{NO},\) and 0.96 mole of \(\mathrm{NO}_{2} .\) Calculate \(K_{P}\) for the reaction, given that the total pressure is 0.20 atm.

Consider the following equilibrium process at \(700^{\circ} \mathrm{C}\) $$ 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{~S}(g) $$ Analysis shows that there are 2.50 moles of \(\mathrm{H}_{2}, 1.35 \times\) \(10^{-5}\) mole of \(\mathrm{S}_{2}\), and 8.70 moles of \(\mathrm{H}_{2} \mathrm{~S}\) present in a 12.0-L flask at equilibrium. Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the reaction.

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