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For each of the following systems at equilibrium, predict whether the reaction will shift to the right, left, or not be affected by a decrease in temperature. (a) \(\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)\) endothermic (b) \(2 \mathrm{O}_{3}(g) \rightleftharpoons 3 \mathrm{O}_{2}(g)\) exothermic

Short Answer

Expert verified
For system (a), the reaction will shift to the right. For system (b), the reaction will shift to the left.

Step by step solution

01

Understand Le Chatelier's Principle

Le Chatelier's Principle is to maintain the state of equilibrium when an external force is applied. When temperature is decreased, an endothermic reaction (which absorbs heat) will shift to the right, towards the products, to absorb this change (which in this case is a decrease in temperature). A decrease in temperature in an exothermic reactions (which releases heat) will cause the reaction to shift to the left, towards the reactants, to produce more heat to counteract the temperature drop.
02

Apply Le Chatelier's Principle on system (a)

For the reaction \(\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)\), which is endothermic, a decrease in the temperature will cause the reaction to shift to the right, as the reaction would want to counteract this change by absorbing more heat, thus producing more products.
03

Apply Le Chatelier's Principle on system (b)

For the reaction \(2 \mathrm{O}_{3}(g) \rightleftharpoons 3 \mathrm{O}_{2}(g)\), which is exothermic, a decrease in the temperature will cause the reaction to shift to the left, as the reaction would want to counteract this change by releasing more heat, thus producing more reactants.

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Key Concepts

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

Understanding Endothermic Reactions
An endothermic reaction is a type of chemical reaction that requires an input of energy to proceed. These reactions absorb heat from their surroundings, making them feel cooler. Why does this happen? When the reactants transform into products, they need more energy to break apart their bonds than is released when new bonds form. Therefore, these reactions need a continuous supply of energy, usually in the form of heat, to keep going.

Here's another way to think about it: if you imagine the reactants on one side of a hill and the products on the other, endothermic reactions need a push (energy) to climb up the hill. A well-known example of an endothermic process is the melting of ice to water. To melt ice, we need to add heat, which is absorbed and used to change state.

When it comes to equilibrium, Le Chatelier's Principle helps us predict what will happen if changes occur. If a reaction is endothermic and the temperature decreases, the system will try to absorb more heat, resulting in a shift towards the products, or to the right.
  • Absorbs heat from surroundings.
  • Feels cool to the touch due to heat absorption.
  • Shifts right when temperature drops, according to Le Chatelier's Principle.
Exploring Exothermic Reactions
Exothermic reactions are the opposite of endothermic ones. In exothermic reactions, energy is released into the surroundings, usually in the form of heat. These reactions often feel warm or hot to the touch, as the energy given off is often thermal.

During these reactions, the energy needed to break the bonds of reactants is less than the energy released when new bonds form in the products. Imagine coasting downhill - the reaction spontaneously releases energy when transitioning from reactants to products. Combustion, such as burning wood or fuel, is a classic example of an exothermic process, releasing heat and light.

Applying Le Chatelier's Principle, if the temperature lowers in an exothermic reaction, the equilibrium will shift towards the reactants, or to the left, as the system releases additional heat to counteract the temperature drop.
  • Releases heat into surroundings.
  • Feels warm or hot due to energy release.
  • Shifts left when temperature is lowered, based on Le Chatelier's Principle.
Chemical Equilibrium and Le Chatelier's Principle
Chemical equilibrium is a fascinating state where the forward and reverse reactions occur at the same rate, so there is no net change in the concentration of reactants and products. This doesn't mean the reactions have stopped; rather, they're occurring continuously and dynamically.

At equilibrium, the mixture's composition remains constant. Remarkably, by applying stress such as changes in temperature, pressure, or concentration to a system at equilibrium, we can influence its position. This is where Le Chatelier's Principle comes into play.

Le Chatelier's Principle is a handy tool that predicts how a change in conditions affects a system at equilibrium. It states that if the external conditions change, the equilibrium will shift to offset this change and restore balance. For instance, if temperature changes, an endothermic or exothermic reaction will shift to counteract that change - seeking to absorb extra energy or release excess heat as needed.
  • Forward and reverse reactions balance each other out.
  • Le Chatelier's Principle predicts the effect of external changes.
  • Helpful for understanding and manipulating reaction outcomes.

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Most popular questions from this chapter

Consider the following reaction and its equilibrium constant at \(100^{\circ} \mathrm{C}\) : $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \quad K_{\mathrm{eq}}=6.5 $$ If \(0.250 \mathrm{~mol}\) of each reactant and product is mixed into a \(1.0\) - \(\mathrm{L}\) container, will the reaction proceed in the forward or reverse direction, or is it already at equilibrium?

Consider the following exothermic reaction: $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ Which of the following changes will increase the number of moles of NO at equilibrium? Explain why or why not for each change. (a) remove \(\mathrm{H}_{2} \mathrm{O}\) (b) decrease volume (c) decrease temperature (d) add \(\mathrm{O}_{2}\) (e) add a catalyst

Consider the following two-step reaction: $$ \begin{aligned} \mathrm{Cu}^{2+}+\mathrm{H}_{2} & \longrightarrow \mathrm{CuH}^{+}+\mathrm{H}^{+} \\ \mathrm{CuH}^{+}+\mathrm{H}^{+}+\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2} & \longrightarrow \mathrm{Cu}^{2+}+\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{3} \end{aligned} $$ (a) Identify any catalysts or intermediates. (b) Write the net reaction.

Are all collisions between reactants effective? Why or why not?

If the initial concentrations of reactants and products are substituted into the equilibrium constant expression, and the value obtained is greater than the equilibrium constant, is the system in a state of equilibrium? If not, in which direction will the reaction shift to reach equilibrium? Explain.

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