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Chapter 6: Problem 27

Consider the reaction $$\begin{array}{r}2 \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \\ \Delta H=483.6 \mathrm{~kJ} / \mathrm{mol}\end{array}$$ If 2.0 moles of \(\mathrm{H}_{2} \mathrm{O}(g)\) are converted to \(\mathrm{H}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) against a pressure of \(1.0 \mathrm{~atm}\) at \(125^{\circ} \mathrm{C},\) what is \(\Delta U\) for this reaction?

Short Answer

Expert verified
\(\Delta U\) for this reaction is approximately 483275.84 J/mol.

Step by step solution

01

Recall the First law of Thermodynamics

The first law of thermodynamics can be represented as \( \Delta U = q + w \), where \( \Delta U \) is the change in internal energy, \( q \) is the heat transferred and \( w \) is the work done on the system. In this case, we can consider the reaction to be at constant pressure, leading to the formula \( \Delta H = \Delta U + P\Delta V \), where \( P\Delta V \) describes the work done on the system.
02

Calculate the change in volume (\( \Delta V \))

Because the ideal gas law can be written as \(PV=nRT\), we can express the change in volume with \(\Delta V = nRT/P \). However, we need to use the given temperature in Kelvin. So, convert the temperature from Celsius to Kelvin by adding 273 to the temperature in Celsius, which gives 125°C + 273 = 398 K. Plugging the numbers into the formula yields, \(\Delta V = 3.2 L \).
03

Calculate the work done (\( P\Delta V \))

Here, the work done can be calculated using the formula \(P\Delta V\), where \(P\) is the pressure and \(\Delta V\) is the change in volume. We know pressure is 1.0 atm which is equals to 101.3 J/(mol.L) and volume change \(\Delta V = 3.2 L\). Therefore, \(P\Delta V\) equals 324.16 J/mol.
04

Find \(\Delta U\)

Finally, we can solve for \(\Delta U\) using the first law of thermodynamics. The formula is \(\Delta H = \Delta U + P\Delta V\). We know \(\Delta H = 483.6 kJ/mol = 483600 J/mol\) and the work done \(P\Delta V = 324.16 J/mol\). Rearranging the formula and substituting the known values gives \(\Delta U = \Delta H - P\Delta V = 483600 J/mol - 324.16 J/mol = 483275.84 J/mol\).

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Most popular questions from this chapter

In writing thermochemical equations, why is it important to indicate the physical state (that is, gaseous, liquid, solid, or aqueous) of each substance?

A quantity of \(2.00 \times 10^{2} \mathrm{~mL}\) of \(0.862 \mathrm{MHCl}\) is mixed with an equal volume of \(0.431 M \mathrm{Ba}(\mathrm{OH})_{2}\) in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the \(\mathrm{HCl}\) and \(\mathrm{Ba}(\mathrm{OH})_{2}\) solutions is the same at \(20.48^{\circ} \mathrm{C}\), For the process $$\mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)$$ the heat of neutralization is \(-56.2 \mathrm{~kJ} / \mathrm{mol}\). What is the final temperature of the mixed solution?

A 3.53-g sample of ammonium nitrate \(\left(\mathrm{NH}_{4} \mathrm{NO}_{3}\right)\) was added to \(80.0 \mathrm{~mL}\) of water in a constantpressure calorimeter of negligible heat capacity. As a result, the temperature of the water decreased from \(21.6^{\circ} \mathrm{C}\) to \(18.1^{\circ} \mathrm{C}\). Calculate the heat of solution \(\left(\Delta H_{\mathrm{soln}}\right)\) of ammonium nitrate.

The standard enthalpies of formation of ions in aqueous solutions are obtained by arbitrarily assigning a value of zero to \(\mathrm{H}^{+}\) ions; that is, \(\Delta H_{\mathrm{f}}^{\circ}\left[\mathrm{H}^{+}(a q)\right]=0\) (a) For the following reaction $$\begin{array}{r}\mathrm{HCl}(g) \stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{H}^{+}(aq)+\mathrm{Cl}^{-}(a q) \\\\\Delta H^{\circ}=-74.9 \mathrm{~kJ} / \mathrm{mol}\end{array}$$ calculate \(\Delta H_{\mathrm{f}}^{\circ}\) for the \(\mathrm{Cl}^{-}\) ions. (b) Given that \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{OH}^{-}\) ions is \(-229.6 \mathrm{~kJ} / \mathrm{mol}\), calculate the enthalpy of neutralization when 1 mole of a strong monoprotic acid (such as \(\mathrm{HCl}\) ) is titrated by 1 mole of a strong base (such as \(\mathrm{KOH}\) ) at \(25^{\circ} \mathrm{C}\).

Explain the meaning of this thermochemical equation: $$\begin{array}{r}4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\\\\Delta H=-904 \mathrm{~kJ} / \mathrm{mol}\end{array}$$
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