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Consider this reaction: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ The equilibrium constant \(K_{P}\) for the reaction is \(1.0 \times\) \(10^{-15}\) at \(25^{\circ} \mathrm{C}\) and 0.050 at \(2200^{\circ} \mathrm{C}\). Is the formation of nitric oxide endothermic or exothermic? Explain your answer.

Short Answer

Expert verified
The formation of nitric oxide is endothermic since an increase in temperature favors the formation of nitric oxide, implying heat is absorbed in the reaction.

Step by step solution

01

Understand the terms endothermic and exothermic

Endothermic reactions are those where energy (in the form of heat) is absorbed from the surroundings, causing the reaction to feel cold. Exothermic reactions are those where energy is released into the surroundings, causing the reaction to feel hot.
02

Interpret the given equilibrium constants

The value of the equilibrium constant \(K_P\) increases from \(1.0 \times 10^{-15}\) at \(25^{\circ} C\) to 0.050 at \(2200^{\circ} C\). This shows that as the temperature increases, the equilibrium of the reaction shifts to the right, favoring the formation of nitric oxide (NO).
03

Apply Van't Hoff's Equation

Van't Hoff’s Equation is given by: \(\ln \left(K_2/K_1\right) = -\Delta H/R\left(1/T_2 - 1/T_1\right)\), where \(K_1\) and \(K_2\) are equilibrium constants at temperatures \(T_1\) and \(T_2\) respectively, \(\Delta H\) is the change in enthalpy, and \(R\) is the universal gas constant. If the equilibrium constant increases with increase in temperature, the reaction is endothermic and enthalpy change \(\Delta H\) is positive.
04

Conclude and explain

Therefore, the formation of nitric oxide is endothermic because the reaction absorbs heat to proceed. As the temperature increases, the reaction is favored, which is consistent with endothermic reactions. This fits with the Le Chatelier’s principle, which states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.

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Key Concepts

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

Equilibrium Constant
The equilibrium constant, denoted as \(K\), is a value that indicates the ratio of concentrations of products to reactants at equilibrium for a given chemical reaction at a specific temperature. It is a critical concept in understanding how a reaction proceeds and reaches a state where the rates of forward and reverse reactions are equal, leading to no net change in the concentration of reactants and products over time.

An equilibrium constant can be expressed in terms of partial pressures (\(K_P\)) or concentrations (\(K_C\)), and its value can provide insights into whether products or reactants are favored at equilibrium. A small \(K\) value (much less than 1) suggests the reaction favors the reactants, whereas a large \(K\) value (much greater than 1) suggests the reaction favors the products. By analyzing the change in the equilibrium constant with temperature, we can infer the reaction's heat dependence. In the exercise, \(K_P\) increases significantly with an increase in temperature, indicating that the formation of products (in this case, nitric oxide) is favored at higher temperatures.
Van't Hoff's Equation
Van't Hoff's Equation is a mathematical description of how the equilibrium constant changes with temperature. It is an important equation in chemical kinetics and thermodynamics. The equation is given as \(\ln \left(K_2/K_1\right) = -\Delta H/R\left(1/T_2 - 1/T_1\right)\), where \(K_1\) and \(K_2\) are the equilibrium constants at temperatures \(T_1\) and \(T_2\), respectively, \(\Delta H\) is the enthalpy change of the reaction, \(R\) is the universal gas constant, and \(T\) represents the temperatures in Kelvins.

According to this equation, if the equilibrium constant increases with an increase in temperature (as observed in the exercise for the formation of nitric oxide), the reaction is likely to be endothermic, because \(\Delta H\) will be positive. Conversely, if the equilibrium constant decreases, the reaction would be exothermic with a negative \(\Delta H\). This equation is vital for predicting the direction and extent of a chemical reaction under varying temperatures.
Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept in chemistry that describes how a system at equilibrium reacts to disturbances. It states that if a dynamic equilibrium is disturbed by changing the conditions, such as temperature, pressure, or concentration, the position of equilibrium moves to counteract the change.

For example, increasing the temperature of an endothermic reaction will shift the equilibrium towards the product side to absorb the added heat, thereby partially counteracting the temperature change. Conversely, for an exothermic reaction, increasing the temperature will shift equilibrium towards the reactants. In the context of the given exercise, the principle helps explain why the equilibrium constant increases with temperature for the reaction: as the heat is introduced (temperature rises), the equilibrium shifts to favor the formation of nitric oxide, which absorbs this heat, indicating an endothermic process.
Enthalpy Change
Enthalpy change, represented by \(\Delta H\), is a measure of the total heat content of a system and reflects the energy absorbed or released during a chemical reaction. It is an essential concept for understanding thermodynamics and reaction energetics. A positive \(\Delta H\) signifies that the reaction is endothermic, absorbing heat from its surroundings. In contrast, a negative \(\Delta H\) means the reaction is exothermic, releasing heat into its surroundings.

Knowing the enthalpy change allows us to predict the effect of temperature on a reaction's equilibrium. As illustrated in the exercise where the \(K_P\) for nitric oxide increases with temperature, we can deduce that the reaction's enthalpy change is positive. This correlates with the observation that producing nitric oxide requires heat absorption, making it an endothermic reaction.

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Most popular questions from this chapter

Consider this reaction: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ If the equilibrium partial pressures of \(\mathrm{N}_{2}, \mathrm{O}_{2}\), and NO are 0.15 atm, 0.33 atm, and 0.050 atm, respectively, at \(2200^{\circ} \mathrm{C}\), what is \(K_{P} ?\)

About 75 percent of hydrogen for industrial use is produced by the steam- reforming process. This process is carried out in two stages called primary and secondary reforming. In the primary stage, a mixture of steam and methane at about 30 atm is heated over a nickel catalyst at \(800^{\circ} \mathrm{C}\) to give hydrogen and carbon monoxide: $$ \begin{aligned} \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g) &+3 \mathrm{H}_{2}(g) \\ \Delta H^{\circ} &=206 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ The secondary stage is carried out at about \(1000^{\circ} \mathrm{C}\) in the presence of air, to convert the remaining methane to hydrogen: \(\mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+2 \mathrm{H}_{2}(g)\) $$ \Delta H^{\circ}=35.7 \mathrm{~kJ} / \mathrm{mol} $$ (a) What conditions of temperature and pressure would favor the formation of products in both the primary and secondary stages? (b) The equilibrium constant \(K_{\mathrm{c}}\) for the primary stage is 18 at \(800^{\circ} \mathrm{C}\). (i) Calculate \(K_{P}\) for the reaction. (ii) If the partial pressures of methane and steam were both 15 atm at the start, what are the pressures of all the gases at equilibrium?

What effect does an increase in pressure have on each of these systems at equilibrium? (a) \(\mathrm{A}(s) \rightleftharpoons 2 \mathrm{~B}(s)\) (b) \(2 \mathrm{~A}(l) \rightleftharpoons \mathrm{B}(l)\) (c) \(\mathrm{A}(s) \rightleftharpoons \mathrm{B}(g)\) (d) \(\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g)\) (e) \(\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g)\) The temperature is kept constant. In each case, the reacting mixture is in a cylinder fitted with a movable piston.

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 \(298 \mathrm{~K}\).

What is meant by "the position of an equilibrium"? Does the addition of a catalyst have any effects on the position of an equilibrium?

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