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Chapter 6: Problem 50

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. \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) 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: - Reaction a (\(2 \mathrm{HF}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g)\)): \(\Delta H = \Delta E\). - Reaction b (\(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\)): \(\Delta H<\Delta E\). - Reaction c (\(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\)): \(\Delta H>\Delta E\).

Step by step solution

01

Reaction a

Determine the change in the number of moles for reaction a: \(2 \mathrm{HF}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g)\). On the left side, there are 2 moles of \(\mathrm{HF}(g)\), and on the right, there is 1 mole of \(\mathrm{H}_{2}(g)\) and 1 mole of \(\mathrm{F}_{2}(g)\). Therefore, \(\Delta n =1+1-2=0\).
02

Reaction a Comparisons

Because \(\Delta n=0\) for reaction a, there is no change in the number of moles, and \(\Delta V=0\). Thus, \(\Delta H=\Delta E+P \Delta V=\Delta E\).
03

Reaction b

Determine the change in the number of moles for reaction b: \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\). On the left side, there are 1 mole of \(\mathrm{N}_{2}(g)\) and 3 moles of \(\mathrm{H}_{2}(g)\), and on the right, there are 2 moles of \(\mathrm{NH}_{3}(g)\). Therefore, \(\Delta n = 2 - (1+3)=-2\).
04

Reaction b Comparisons

Since \(\Delta n=-2\) for reaction b, there is a decrease in the number of moles, and \(\Delta V<0\). Thus, \(\Delta H=\Delta E+P \Delta V<\Delta E\).
05

Reaction c

Determine the change in the number of moles for reaction c: \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\). On the left side, there are 4 moles of \(\mathrm{NH}_{3}(g)\) and 5 moles of \(\mathrm{O}_{2}(g)\), and on the right, there are 4 moles of \(\mathrm{NO}(g)\) and 6 moles of \(\mathrm{H}_{2}\mathrm{O}(g)\). Therefore, \(\Delta n = (4+6)-(4+5)=1\).
06

Reaction c Comparisons

Because \(\Delta n=1\) for reaction c, there is an increase in the number of moles, and \(\Delta V>0\). Thus, \(\Delta H=\Delta E+P \Delta V>\Delta E\). In summary: - For reaction a, \(\Delta H = \Delta E\). - For reaction b, \(\Delta H<\Delta E\). - For reaction c, \(\Delta H>\Delta E\).

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Most popular questions from this chapter

A \(30.0\) -g sample of water at \(280 . \mathrm{K}\) is mixed with \(50.0 \mathrm{~g}\) water at \(330 . \mathrm{K}\). Calculate the final temperature of the mixture assuming no heat loss to the surroundings.

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For the process \(\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)\) at \(298 \mathrm{~K}\) and \(1.0 \mathrm{~atm}\), \(\Delta H\) is more positive than \(\Delta E\) by \(2.5 \mathrm{~kJ} / \mathrm{mol}\). What does the \(2.5 \mathrm{~kJ} / \mathrm{mol}\) quantity represent?

If the internal energy of a thermodynamic system is increased by \(300 . \mathrm{J}\) while \(75 \mathrm{~J}\) of expansion work is done, how much heat was transferred and in which direction, to or from the system?

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 \(150.0 \mathrm{~g}\) Ag from \(273 \mathrm{~K}\) to \(298 \mathrm{~K}\). b. Calculate the energy required to raise the temperature of \(1.0\) mole of 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.
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