Predict the effect of decreasing the temperature on the amount(s) of reactant(s) in the following reactions: (a) \(\mathrm{C}_{2} \mathrm{H}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{CHO}(g) \quad \Delta H_{\mathrm{rxn}}^{\circ}=-151 \mathrm{~kJ}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(l)+\mathrm{H}_{2} \mathrm{O}(g)\) $$ \Delta H_{\mathrm{rxn}}^{\circ}=-451 \mathrm{~kJ} $$ (c) \(2 \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CH}_{3} \mathrm{CHO}(g)\) (exothermic) (d) \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) (endothermic)

Short Answer

Expert verified
For reactions (a), (b), and (c), decreasing the temperature decreases the amount of reactants. For reaction (d), decreasing the temperature increases the amount of reactant.

Step by step solution

01

Understand Le Chatelier's Principle

Le Chatelier's Principle states that if a system at equilibrium is subjected to a change in temperature, pressure, or concentration, the system will adjust itself to partially counteract that change. Specifically, for temperature changes: if you decrease the temperature, the equilibrium will shift in the direction that produces heat (exothermic reaction).
02

Analyze Reaction (a)

The reaction is \(\mathrm{C}_{2} \mathrm{H}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{CHO}(g) \) with \( \Delta H_{\mathrm{rxn}} = -151 \mathrm{~kJ} \). Since the reaction is exothermic (negative \(\Delta H\)), decreasing the temperature will shift the equilibrium to the right, increasing the amount of products and decreasing the amount of reactants.
03

Analyze Reaction (b)

The reaction is \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(l)+\mathrm{H}_{2} \mathrm{O}(g) \) with \( \Delta H_{\mathrm{rxn}} = -451 \mathrm{~kJ} \). Since the reaction is exothermic, decreasing the temperature will shift the equilibrium to the right, increasing the amount of products and decreasing the amount of reactants.
04

Analyze Reaction (c)

The reaction is \(2 \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CH}_{3} \mathrm{CHO}(g) \) and it is given as exothermic. Decreasing the temperature will again shift the equilibrium to the right, increasing the amount of products and decreasing the amount of reactants.
05

Analyze Reaction (d)

The reaction is \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \) and it is endothermic (requires heat). Decreasing the temperature will shift the equilibrium to the left, increasing the amount of \(\mathrm{N}_{2} \mathrm{O}_{4} \) and decreasing the amount of \(\mathrm{NO}_{2} \).

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Key Concepts

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

equilibrium shift
One of the critical concepts in chemistry is the idea of an equilibrium shift. This occurs when a system in equilibrium responds to a change in temperature, pressure, or concentration of reactants or products. According to Le Chatelier's Principle, if such a change is introduced, the system will adjust to oppose the effect of the change.
For instance, if the concentration of a reactant is increased, the system shifts to produce more products to balance the added reactant. Conversely, if the temperature is changed, the system will shift in a direction that counteracts the temperature change.
To understand this better, consider a simple reversible reaction:
\(\text{A + B} \rightleftharpoons \text{C + D} \)
If the temperature of this exothermic reaction decreases, the equilibrium will shift to the right, favoring the production of C and D. By shifting the balance, the system compensates for the lost heat, illustrating an equilibrium shift.
exothermic reaction
An exothermic reaction is a chemical reaction that releases heat into its surroundings. In such reactions, the energy of the products is lower than the energy of the reactants. This release of energy leads to a negative change in enthalpy (\( \Delta H \)) for the reaction.
For example, consider the reaction:
\(\text{C}_{2}\text{H}_{2}(g) + \text{H}_{2}\text{O}(g) \rightleftharpoons \text{CH}_{3}\text{CHO}(g) \) \( \Delta H = -151 kJ \)
The negative \( \Delta H \) indicates that this reaction is exothermic. Decreasing the temperature of such a reaction will shift the equilibrium to the right, increasing the production of products and reducing reactants. This shift happens because the system tries to counteract the loss in temperature by producing more heat.
endothermic reaction
In contrast to exothermic reactions, endothermic reactions absorb heat from their surroundings. The energy needed to form the products is higher than the energy of the reactants, resulting in a positive change in enthalpy (\( \Delta H \)).
For instance, consider the reaction:
\(\text{N}_{2}\text{O}_{4}(g) \rightleftharpoons 2 \text{NO}_{2}(g) \) (endothermic)
Since this reaction requires heat, decreasing the temperature will shift the equilibrium to the left, increasing the concentration of \( \text{N}_{2}\text{O}_{4} \) and decreasing the concentration of \( \text{NO}_{2} \).
This shift happens because the system compensates for the reduced temperature by favoring the reaction that absorbs heat. Thus, understanding endothermic reactions is essential for predicting how temperature changes impact equilibrium states in chemical reactions.

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Most popular questions from this chapter

For the following equilibrium system, which of the changes will form more \(\mathrm{CaCO}_{3} ?\) $$ \begin{array}{r} \mathrm{CO}_{2}(g)+\mathrm{Ca}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{CaCO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H^{\circ}=-113 \mathrm{~kJ} \end{array} $$ (a) Decrease temperature at constant pressure (no phase change). (b) Increase volume at constant temperature. (c) Increase partial pressure of \(\mathrm{CO}_{2}\). (d) Remove one-half of the initial \(\mathrm{CaCO}_{3}\).

In the \(1980 \mathrm{~s}, \mathrm{CFC}-11\) was one of the most heavily produced chlorofluorocarbons. The last step in its formation is $$ \mathrm{CCl}_{4}(g)+\mathrm{HF}(g) \rightleftharpoons \mathrm{CFCl}_{3}(g)+\mathrm{HCl}(g) $$ If you start the reaction with equal concentrations of \(\mathrm{CCl}_{4}\) and \(\mathrm{HF}\), you obtain equal concentrations of \(\mathrm{CFCl}_{3}\) and \(\mathrm{HCl}\) at equilibrium. Are the final concentrations of \(\mathrm{CFCl}_{3}\) and \(\mathrm{HCl}\) equal if you start with unequal concentrations of \(\mathrm{CCl}_{4}\) and HF? Explain.

When \(0.100 \mathrm{~mol}\) of \(\mathrm{CaCO}_{3}(s)\) and \(0.100 \mathrm{~mol}\) of \(\mathrm{CaO}(s)\) are placed in an evacuated, sealed 10.0 -L container and heated to \(385 \mathrm{~K}, P_{\mathrm{CO}_{2}}=0.220 \mathrm{~atm}\) after equilibrium is established: $$ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ An additional 0.300 atm of \(\mathrm{CO}_{2}(g)\) is pumped in. What is the total mass (in \(\mathrm{g}\) ) of \(\mathrm{CaCO}_{3}\) after equilibrium is re- established?

The two most abundant atmospheric gases react to a tiny extent at \(298 \mathrm{~K}\) in the presence of a catalyst: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) \quad K_{\mathrm{p}}=4.35 \times 10^{-31} $$ (a) What are the equilibrium pressures of the three gases when the atmospheric partial pressures of \(\mathrm{O}_{2}(0.210 \mathrm{~atm})\) and of \(\mathrm{N}_{2}(0.780 \mathrm{~atm})\) are put into an evacuated \(1.00-\mathrm{L}\) flask at \(298 \mathrm{~K}\) with the catalyst? (b) What is \(P_{\text {total }}\) in the container? (c) Find \(K_{\text {c }}\) at \(298 \mathrm{~K}\).

An important industrial source of ethanol is the reaction, catalyzed by \(\mathrm{H}_{3} \mathrm{PO}_{4},\) of steam with ethylene derived from oil: $$ \begin{array}{c} \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-47.8 \mathrm{~kJ} \quad K_{\mathrm{c}}=9 \times 10^{3} \mathrm{at} 600 . \mathrm{K} \end{array} $$ (a) At equilibrium, \(P_{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}=200 . \mathrm{atm}\) and \(P_{\mathrm{H}_{0} \mathrm{O}}=400 .\) atm. Cal- culate \(P_{\mathrm{C}_{2} \mathrm{H}_{4}}\). (b) Is the highest yield of ethanol obtained at high or low \(P\) ? High or low \(T ?\) (c) Calculate \(K_{c}\) at \(450 .\) K. (d) In \(\mathrm{NH}_{3}\) manufacture, the yield is increased by condensing the \(\mathrm{NH}_{3}\) to a liquid and removing it. Would condensing the \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) have the same effect in ethanol production? Explain.

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