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Chapter 15: Problem 65

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 \(\mathrm{C}\) to calculate \(\Delta H^{\circ}\) for this reaction. (b) Will the equilibrium constant for the reaction increase or decrease with increasing temperature? Explain. (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 for the given reaction is -319.8 kJ/mol, indicating that it is an exothermic process. As temperature increases, the equilibrium constant, K, will decrease since the system shifts towards the endothermic (reverse) reaction. At constant temperature, increasing the volume will decrease the fraction of products in the equilibrium mixture, as the equilibrium shifts towards the side with more moles of gas (reactants). Conversely, decreasing the volume will increase the fraction of products, as the equilibrium shifts towards the side with fewer moles of gas (products).

Step by step solution

01

Find the standard enthalpy change of each reactant and product

To calculate the standard enthalpy change for the reaction, we need to find the enthalpy change of each reactant and product. We are required to use data from Appendix C (not provided here) to find the values. We assume that you have access to the standard enthalpy changes of formation. Suppose the values are: Standard enthalpy change of formation for NO (g): 90.3 kJ/mol Standard enthalpy change of formation for NO₂ (g): 33.2 kJ/mol Standard enthalpy change of formation for N₂O (g): -82.1 kJ/mol
02

Apply Hess's Law to calculate the standard enthalpy change of the reaction

We can use Hess's Law to calculate the ΔH° for the given reaction. Hess's Law states that the enthalpy change of the reaction is equal to the sum of the enthalpy changes of products minus the sum of the enthalpy changes of reactants. For this reaction, we have: ΔH° = ([ΔH°(NO₂) + ΔH°(N₂O)] - [3 × ΔH°(NO)]) Using the values found in step 1: ΔH° = [(33.2 kJ/mol) + (-82.1 kJ/mol)] - [3 × (90.3 kJ/mol)]
03

Calculate the standard enthalpy change of the reaction

Now, we can calculate the ΔH° for the reaction: ΔH° = (-48.9 kJ/mol) - (270.9 kJ/mol) ΔH° = -319.8 kJ/mol So, the standard enthalpy change for this reaction is -319.8 kJ/mol. #Part(b)#
04

Determine the sign of ΔH° and its effect on equilibrium

In part (a), we calculated the ΔH° as -319.8 kJ/mol. This negative value indicates that the reaction is exothermic (releases heat). According to Le Chatelier's principle, if the temperature increases, the equilibrium will shift in the direction that absorbs the heat, in this case, the reverse reaction (endothermic).
05

Analyze the effect on the equilibrium constant K

Since increasing temperature causes the equilibrium to shift towards the endothermic (reverse) reaction, the equilibrium constant K will decrease with increasing temperature. #Part(c)#
06

Determine the effect of volume changes on pressure

According to the ideal gas law (PV = nRT), an increase in volume at constant temperature and amount of gas will decrease the pressure, and vice versa.
07

Apply Le Chatelier's principle to analyze the effect of volume changes

Le Chatelier's principle states that a change in a factor, such as pressure, will cause the equilibrium to shift in the direction that counteracts that change. In the given reaction, there are 3 moles of reactant (3NO) and 2 moles of products (NO₂ + N₂O). Therefore, increasing the volume (decreasing the pressure) will shift the equilibrium towards the side with more moles of gas, i.e., towards reactants. Similarly, decreasing the volume (increasing the pressure) will shift the equilibrium towards the side with fewer moles of gas, i.e., towards the products.
08

Determine the effect of volume changes on the fraction of products

The fraction of products in the equilibrium mixture will increase when the volume is decreased (pressure increased), as the equilibrium will tend to shift towards the side with fewer moles of gas. Conversely, the fraction of products will decrease when the volume is increased (pressure decreased), as the equilibrium shifts towards the side with more moles of gas.

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Most popular questions from this chapter

The protein hemoglobin (Hb) transports \(\mathrm{O}_{2}\) in mammalian blood. Each \(\mathrm{Hb}\) can bind \(4 \mathrm{O}_{2}\) molecules. The equilibrium constant for the \(\mathrm{O}_{2}\) -binding reaction is higher in fetal hemoglobin than in adult hemoglobin. In discussing protein oxygenbinding capacity, biochemists use a measure called the \(P 50\) value, defined as the partial pressure of oxygen at which \(50 \%\) of the protein is saturated. Fetal hemoglobin has a \(\mathrm{P} 50\) value of 19 torr, and adult hemoglobin has a P50 value of 26.8 torr. Use these data to estimate how much larger \(K_{c}\) is for the aqueous reaction \(4 \mathrm{O}_{2}(g)+\mathrm{Hb}(a q) \longrightarrow\left[\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)\right]\) .

Write the expression for \(K_{c}\) for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous. (a) \(3 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g)\) (b) \(\mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g)\) (c) \(\mathrm{Ni}(\mathrm{CO})_{4}(g) \rightleftharpoons \mathrm{Ni}(s)+4 \mathrm{CO}(g)\) (d) \(\mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q)\) (e) \(2 \mathrm{Ag}(s)+\mathrm{Zn}^{2+}(a q) \rightleftharpoons 2 \mathrm{Ag}^{+}(a q)+\mathrm{Zn}(s)\) (f) \(\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)\) (g) \(2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons 2 \mathrm{H}^{+}(a q)+2 \mathrm{OH}^{-}(a q)\)

Write the expressions for \(K_{c}\) for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous. (a) \(2 \mathrm{O}_{3}(g) \rightleftharpoons 3 \mathrm{O}_{2}(g)\) (b) \(\mathrm{Ti}(s)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{TiCl}_{4}(l)\) (c) \(2 \mathrm{C}_{2} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g)\) (d) \(\mathrm{C}(s)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)\) (e) \(4 \mathrm{HCl}(a q)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{Cl}_{2}(g)\) (f) \(2 \mathrm{C}_{8} \mathrm{H}_{18}(l)+25 \mathrm{O}_{2}(g) \rightleftharpoons 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(g)\) (g) \(2 \mathrm{C}_{8} \mathrm{H}_{18}(l)+25 \mathrm{O}_{2}(g) \rightleftharpoons 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(l)\)

Both the forward reaction and the reverse reaction in the following equilibrium are believed to be elementary steps: $$ \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}(g)+\mathrm{Cl}(g) $$ At \(25^{\circ} \mathrm{C}\) the rate constants for the forward and reverse reactions are \(1.4 \times 10^{-28} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) and \(9.3 \times 10^{10} \mathrm{M}^{-1} \mathrm{~s}^{-1}\), respectively. (a) What is the value for the equilibrium constant at \(25^{\circ} \mathrm{C} ?\) (b) Are reactants or products more plentiful at equilibrium? (c) What additional information would you need in order to decide whether the reaction as written is endothermic or exothermic?

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) can be made by the reaction of \(\mathrm{CO}\) with \(\mathrm{H}_{2}\) : $$\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g)$$ (a) Use thermochemical data in Appendix \(\mathrm{C}\) to calculate \(\Delta H^{\circ}\) for this reaction. (b) To maximize the equilibrium yield of methanol, would you use a high or low temperature? (c) To maximize the equilibrium yield of methanol, would you use a high or low pressure?
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