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Chapter 6: Problem 54

For the following reactions at constant pressure, predict if $\Delta H>\Delta E, \Delta H<\Delta E,\( or \)\Delta H=\Delta E .$ a. \(2 \mathrm{HF}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g)\) b. $\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)$ c. $4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$

Short Answer

Expert verified
For the given reactions, we have the following relationships between ΔH and ΔE: a. In the reaction \(2 \mathrm{HF}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g)\), the number of moles of gaseous particles remains the same. Therefore, \(\Delta H = \Delta E\). b. In the reaction \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\), the number of moles of gaseous particles decreases. Therefore, \(\Delta H < \Delta E\). c. In the reaction \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2}\mathrm{O}(g)\), the number of moles of gaseous particles increases. Therefore, \(\Delta H > \Delta E\).

Step by step solution

01

Count number of moles of gaseous particles

On the reactants side, there are 2 moles of HF(g). On the products side, there's 1 mole each of H₂(g) and F₂(g), summing up to 2 moles. So the number of moles of gaseous particles on both sides is equal.
02

Use the equation ΔH=ΔE+PΔV

Since the number of moles of gaseous particles is the same on both sides, the ΔV(volume change) is zero. Therefore, our equation will look like this: ΔH = ΔE + P(0) ⟹ ΔH = ΔE. In this reaction, \(\Delta H = \Delta E\). b. Predicting the relationship between ΔH and ΔE for reaction: \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\)
03

Count number of moles of gaseous particles

On the reactants side, there are 1 mole of N₂(g) and 3 moles of H₂(g) - total 4 moles. On the products side, there are 2 moles of NH₃(g). So the number of moles of gaseous particles decreases in this reaction.
04

Use the equation ΔH=ΔE+PΔV

Since the number of moles of gaseous particles decreases, the ΔV will be negative. Therefore, our equation will look like this: ΔH = ΔE + P(-ΔV) ⟹ ΔH < ΔE. In this reaction, \(\Delta H < \Delta E\). c. Predicting the relationship between ΔH and ΔE for reaction: $4 NH_{3}(g)+5 O_{2}(g) \longrightarrow 4 NO(g)+6 H_{2} O(g)$
05

Count number of moles of gaseous particles

On the reactants side, there are 4 moles of NH₃(g) and 5 moles of O₂(g) - total 9 moles. On the products side, there are 4 moles of NO(g) and 6 moles of H₂O(g) - total 10 moles. So the number of moles of gaseous particles increases in this reaction.
06

Use the equation ΔH=ΔE+PΔV

Since the number of moles of gaseous particles increases, the ΔV will be positive. Therefore, our equation will look like this: ΔH = ΔE + P(ΔV) ⟹ ΔH > ΔE. In this reaction, \(\Delta H > \Delta E\).

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Most popular questions from this chapter

One mole of \(\mathrm{H}_{2} \mathrm{O}(g)\) at 1.00 atm and $100 .^{\circ} \mathrm{C}\( occupies a volume of 30.6 \)\mathrm{L}$ . When 1 mole of \(\mathrm{H}_{2} \mathrm{O}(g)\) is condensed to 1 mole of \(\mathrm{H}_{2} \mathrm{O}(l)\) at 1.00 atm and $100 .^{\circ} \mathrm{C}, 40.66 \mathrm{kJ}\( of heat is released. If the density of \)\mathrm{H}_{2} \mathrm{O}(l)$ at this temperature and pressure is \(0.996 \mathrm{g} / \mathrm{cm}^{3},\) calculate \(\Delta E\) for the condensation of 1 mole of water at 1.00 \(\mathrm{atm}\) and \(100 .^{\circ} \mathrm{C}\)

Given the following data $$ \begin{array}{ll}{\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=-23 \mathrm{kJ}} \\ {3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=-39 \mathrm{kJ}} \\ {\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}(g) \longrightarrow 3 \mathrm{FeO}(s)+\mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=18 \mathrm{kJ}}\end{array} $$ calculate \(\Delta H^{\circ}\) for the reaction $$ \mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$

The specific heat capacity of silver is 0.24 $\mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}$ a. Calculate the energy required to raise the temperature of 150.0 g Ag from 273 \(\mathrm{K}\) to 298 \(\mathrm{K}\) . b. Calculate the energy required to raise the temperature of 1.0 mole of \(\mathrm{Ag}\) by \(1.0^{\circ} \mathrm{C}\) (called the molar heat capacity of silver). c. It takes 1.25 \(\mathrm{kJ}\) of energy to heat a sample of pure silver from \(12.0^{\circ} \mathrm{C}\) to \(15.2^{\circ} \mathrm{C}\) . Calculate the mass of the sample of silver.

It takes 585 \(\mathrm{J}\) of energy to raise the temperature of 125.6 \(\mathrm{g}\) mercury from \(20.0^{\circ} \mathrm{C}\) to $53.5^{\circ} \mathrm{C}$ . Calculate the specific heat capacity and the molar heat capacity of mercury.

A coffee-cup calorimeter initially contains 125 \(\mathrm{g}\) water at \(24.2^{\circ} \mathrm{C} .\) Potassium bromide \((10.5 \mathrm{g}),\) also at \(24.2^{\circ} \mathrm{C},\) is added to the water, and after the KBr dissolves, the final temperature is \(21.1^{\circ} \mathrm{C}\) . Calculate the enthalpy change for dissolving the salt in \(\mathrm{J} / \mathrm{g}\) and $\mathrm{kJ} / \mathrm{mol}$ . Assume that the specific heat capacity of the solution is 4.18 \(\mathrm{J} / \mathrm{C} \cdot \mathrm{g}\) and that no heat is transferred to the surroundings or to the calorimeter.
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