Consider the following equilibrium between oxides of nitrogen $$3 \mathrm{NO}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g)$$ (a) Use data in Appendix C to calculate \(\Delta H^{\circ}\) for this reaction. (b) Will the equilibrium constant for the reaction increase or decrease with increasing temperature? (c) At constant temperature, would a change in the volume of the container affect the fraction of products in the equilibrium mixture?

Short Answer

Expert verified
The standard enthalpy change (∆H°) for the reaction is -65.3 kJ/mol. Since the reaction is exothermic, the equilibrium constant (K) will decrease with increasing temperature. At constant temperature, increasing the volume will favor the reactants, increasing the fraction of products (NO₂ and N₂O) in the equilibrium mixture. Conversely, decreasing the volume will favor the products, decreasing the fraction of products in the equilibrium mixture.

Step by step solution

01

Find the standard enthalpies of formation

Firstly, we need to find the standard enthalpies of formation (∆H°f) for each gas from Appendix C. The values are as follows: \(∆H°_{f} (NO(g)) = 90.29 \, kJ/mol\) \(∆H°_{f} (NO_{2}(g)) = 33.83 \, kJ/mol\) \(∆H°_{f} (N_{2}O(g)) = 81.57 \, kJ/mol\)
02

Calculate ∆H° for the reaction

Using the standard enthalpies of formation, we can calculate the enthalpy change for the reaction as follows: ∆H° = [ ∆H°f(products) - ∆H°f(reactants) ] ∆H° = [ \(1 \times 33.83 + 1 \times 81.57\) ] - [ \(3 \times 90.29\) ] ∆H° = -65.3 kJ/mol
03

Determine the effect of temperature on equilibrium constant

Next, we need to understand the relationship between temperature and the equilibrium constant. According to Le Chatelier's principle, if the reaction is exothermic (i.e., ∆H° < 0), increasing the temperature will shift the equilibrium towards the side with fewer moles of gas (favoring the reactants in this case). Conversely, if the reaction is endothermic (i.e., ∆H° > 0), increasing the temperature will shift the equilibrium towards the side with more moles of gas (favoring the products in this case). In our case, since ∆H° is negative (-65.3 kJ/mol), the reaction is exothermic, and the equilibrium will shift towards the reactants when the temperature is increased. Therefore, the equilibrium constant (K) will decrease with increasing temperature.
04

Determine the effect of volume change on equilibrium mixture

Lastly, we need to analyze the effect of volume change on the equilibrium mixture at constant temperature. According to Le Chatelier's principle, a decrease in volume increases the pressure, which will favor the side with fewer moles of gas, while an increase in volume reduces the pressure, favoring the side with more moles of gas. In our reaction, the number of moles on the reactant side (3 moles of NO) is greater than the number of moles on the product side (1 mole of NO₂ and 1 mole of N₂O). Thus, at constant temperature, an increase in the volume of the container will shift the equilibrium towards the reactants, increasing the fraction of products in the equilibrium mixture (NO₂ and N₂O). Conversely, a decrease in the container's volume will shift the equilibrium towards the products, decreasing the fraction of products in the equilibrium mixture.

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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 cornerstone concept in chemical equilibrium that helps predict how a system at equilibrium reacts to external changes. Imagine a seesaw perfectly balanced at equilibrium. If we add weight to one side, the seesaw will tilt to restore balance. Similarly, a chemical system at equilibrium will adjust to counteract the imposition of changes such as alterations in concentration, pressure, or temperature.

For example, if we increase the concentration of a reactant, the system shifts to produce more products to reduce the reactants' heightened concentration. Likewise, for pressure, which is akin to crowding, if we reduce the container size (increasing the pressure), the system shifts toward the side with fewer gas particles, essentially seeking more 'elbow room'. Temperature changes are analogous to adding or removing thermal energy: when we heat an equilibrium mixture of an exothermic reaction (releases heat), the system adjusts to absorb excess heat by favoring the reactant side. In the case of our nitrogen oxides reaction, the decrease in temperature would therefore shift the equilibrium toward product formation, conversely to the increase in temperature.
Enthalpy Change Calculation
In thermodynamics, the enthalpy change (abla H) of a reaction signifies the total heat absorbed or released under constant pressure. It’s a measure of the energy change in a system and plays a vital role in determining whether a reaction is endothermic (absorbing heat) or exothermic (releasing heat). To calculate the enthalpy change, you apply the following formula:
abla H = [ abla H^°f(products) - abla H^°f(reactants) ].

This formula hinges on the standard enthalpies of formation (abla H°f), which are the heat changes when one mole of a compound forms from its elements at standard conditions. For reactions including our equilibrium between oxides of nitrogen, we look up these abla H°f values from a reference like Appendix C, then subtract the total abla H°f of the reactants from that of the products. An essential tip to simplify calculations is to remember for elements in their standard states, such as N₂(g), abla H°f is zero. If the result is negative, it signals the reaction is exothermic, releasing heat and likely to decrease in rate with rising temperature.
Equilibrium Constant
The equilibrium constant (K) is a quantitative measure of the proportion of reactants to products at equilibrium. It's derived from the law of mass action, which in its core states the rate of a reaction is proportional to the concentration of the reactants raised to a power equivalent to their coefficients in the balanced equation. For the generalized reaction aA + bB ↔ cC + dD, the equilibrium constant expression is
K = [C]^c [D]^d / [A]^a [B]^b.

The beauty of the equilibrium constant is its invariant nature at a fixed temperature. No matter the starting concentrations, the system will adjust until K is reached. In our nitrogen oxides example, if you changed the temperature, you'd effectively alter the value of K, since it is temperature-dependent. An exothermic reaction (with a negative abla H) will have a K that decreases with increasing temperature, as the system favors the reactant side to dissipate added heat. In contrast, for endothermic reactions, K increases as temperature rises. Understanding K helps in predicting the concentration of reactants and products without knowing the specific dynamics at play.

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Most popular questions from this chapter

For the equilibrium $$\mathrm{PH}_{3} \mathrm{BCl}_{3}(s) \rightleftharpoons \mathrm{PH}_{3}(g)+\mathrm{BCl}_{3}(g)$$ \(K_{p}=0.052\) at \(60^{\circ} \mathrm{C}\) (a) Calculate \(K_{C}\) (b) After 3.00 \(\mathrm{g}\) of solid \(\mathrm{PH}_{3} \mathrm{BCl}_{3}\) is added to a closed 1.500 -L. vessel at \(60^{\circ} \mathrm{C}\) , the vessel is charged with 0.0500 \(\mathrm{g}\) of \(\mathrm{BCl}_{3}(g) .\) What is the equilibrium concentration of \(\mathrm{PH}_{3} ?\)

Consider the equilibrium $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ Calculate the equilibrium constant \(K_{p}\) for this reaction, given the following information (at 298 \(\mathrm{K} )\) : \begin{equation} \begin{array}{l}{2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \operatorname{NOBr}(g) \quad K_{c}=2.0} \\ {2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \quad K_{c}=2.1 \times 10^{30}}\end{array} \end{equation}

Consider the hypothetical reaction \(\mathrm{A}(g)+2 \mathrm{B}(g) \rightleftharpoons\) \(2 \mathrm{C}(g),\) for which \(K_{c}=0.25\) at a certain temperature. A 1.00 -L reaction vessel is loaded with 1.00 mol of compound \(C,\) which is allowed to reach equilibrium. Let the variable \(x\) represent the number of mol/L of compound A present at equilibrium. (a) In terms of \(x,\) what are the equilibrium concentrations of compounds \(\mathrm{B}\) and \(\mathrm{C} ?\) (b) What limits must be placed on the value of \(x\) so that all concentrations are positive? (c) By putting the equilibrium concentrations (in terms of \(x\) ) into the equilibrium-constant expression, derive an equation that can be solved for \(x .(\mathbf{d})\) The equation from part (c) is a cubic equation (one that has the form \(a x^{3}+b x^{2}+c x+d=0 )\) . In general, cubic equations cannot be solved in closed form. However, you can estimate the solution by plotting the cubic equation in the allowed range of \(x\) that you specified in part (b). The point at which the cubic equation crosses the \(x\) -axis is the solution. (e) From the plot in part (d), estimate the equilibrium concentrations of \(A, B,\) and C. (Hint: You can check the accuracy of your answer by substituting these concentrations into the equilibrium expression.)

At \(373 \mathrm{K}, K_{p}=0.416\) for the equilibrium $$2 \operatorname{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ If the pressures of \(\mathrm{NOBr}(g)\) and \(\mathrm{NO}(g)\) are equal, what is the equilibrium pressure of \(\mathrm{Br}_{2}(g) ?\)

If \(K_{c}=1\) for the equilibrium \(2 \mathrm{A}(g) \rightleftharpoons \mathrm{B}(g),\) what is the relationship between \([\mathrm{A}]\) and \([\mathrm{B}]\) at equilibrium?

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