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Chapter 8: Problem 36

Given the following thermochemical equations $$ \begin{aligned} 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & & \Delta H=-571.6 \mathrm{~kJ} \\ \mathrm{~N}_{2} \mathrm{O}_{5}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(l) & & \Delta H=-73.7 \mathrm{~kJ} \\ \frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{O}_{2}(g)+\frac{1}{2} \mathrm{H}_{2}(g) \longrightarrow \mathrm{HNO}_{3}(l) & & \Delta H=-174.1 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta H\) for the formation of one mole of dinitrogen pentoxide from its elements in their stable state at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

Short Answer

Expert verified
The enthalpy change for the formation of one mole of dinitrogen pentoxide from its elements in their stable state at 25°C and 1 atm is -1067.2 kJ.

Step by step solution

01

1. Identify the target reaction

We want to find the enthalpy change for the formation of N₂O₅ from its elements, which means the target reaction is: $$ \frac{1}{2} N_{2}(g) + \frac{5}{2} O_{2}(g) \longrightarrow N_{2}O_{5}(g) $$
02

2. Manipulate the given equations as needed

We'll need to flip the second equation and multiply it by 2, because we want to go from 2 HNO₃ to N₂O₅ + H₂O. For the third equation, we can flip it and multiply by 2 as well. The manipulated equations become: $$ \begin{aligned} 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & & \Delta H=-571.6 \mathrm{~kJ} \\ 2 \mathrm{HNO}_{3}(l) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{5}(g)+\mathrm{H}_{2} \mathrm{O}(l) & & \Delta H=2(-73.7 \mathrm{~kJ}) \\ \mathrm{HNO}_{3}(l) \longrightarrow \frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{O}_{2}(g)+\frac{1}{2} \mathrm{H}_{2}(g) & & \Delta H=2(-174.1 \mathrm{~kJ}) \end{aligned} $$
03

3. Add the relevant equations

We now add the three manipulated equations: Equation 1: $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \qquad\qquad\Delta H=-571.6 \,\text{kJ} $$ Equation 2: $$ 2 \mathrm{HNO}_{3}(l) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{5}(g)+\mathrm{H}_{2} \mathrm{O}(l)\qquad\Delta H=2(-73.7) \,\text{kJ} $$ Equation 3: $$ \mathrm{HNO}_{3}(l) \longrightarrow \frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{O}_{2}(g)+\frac{1}{2}\mathrm{H}_{2}(g) \qquad\qquad\Delta H=2(-174.1)\,\text{kJ} $$ Sum of the equations: $$ \frac{1}{2} N_{2}(g) + \frac{5}{2} O_{2}(g) \longrightarrow N_{2}O_{5}(g) \qquad\qquad\Delta H = ? $$
04

4. Calculate the enthalpy change for the target reaction

We'll now use Hess's Law to calculate the enthalpy change for the target reaction, by adding the enthalpy changes of the three manipulated equations: $$ \Delta H_{target} = -571.6 + 2(-73.7) + 2(-174.1) = -571.6 -147.4 - 348.2 = -1067.2 \,\text{kJ} $$ So, the enthalpy change for the formation of one mole of dinitrogen pentoxide from its elements in their stable state at \(25 ^{\circ} \mathrm{C}\) and \(1\,\text{atm}\) is \(-1067.2 \,\text{kJ}\).

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Most popular questions from this chapter

Urea, \(\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}\), is used in the manufacture of resins and glues. When \(5.00 \mathrm{~g}\) of urea is dissolved in \(250.0 \mathrm{~mL}\) of water \((d=1.00 \mathrm{~g} / \mathrm{mL})\) at \(30.0^{\circ} \mathrm{C}\) in a coffee-cup calorimeter, \(27.6 \mathrm{~kJ}\) of heat is absorbed. (a) Is the solution process exothermic? (b) What is \(q_{\mathrm{H}_{2} \mathrm{O}}\) ? (c) What is the final temperature of the solution? (Specific heat of water is \(4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\).) (d) What are the initial and final temperatures in \({ }^{\circ} \mathrm{F}\) ?

Titanium is a metal used in jet engines. Its specific heat is \(0.523 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\). If \(5.88 \mathrm{~g}\) of titanium absorbs \(4.78 \mathrm{~J}\), what is the change in temperature?

How many mL of water at \(10^{\circ} \mathrm{C}\) ( 2 significant figures) must be added to \(75 \mathrm{~mL}\) of water at \(35^{\circ} \mathrm{C}\) to obtain a final temperature of \(19^{\circ} \mathrm{C} ?\) (Make the same assumptions as in Question 9.)

Determine whether the statements given below are true or false. Consider enthalpy \((H)\) (a) It is a state property. (b) \(q_{\text {reaction }}\) (at constant \(\left.P\right)=\Delta H=H_{\text {products }}-H_{\text {reactants }}\) (c) The magnitude of \(\Delta H\) is independent of the amount of reactant. (d) In an exothermic process, the enthalpy of the system remains unchanged.

Consider the following reaction in a vessel with a movable piston. $$ \mathrm{X}(g)+\mathrm{Y}(g) \longrightarrow \mathrm{Z}(l) $$ As the reaction occurs, the system loses \(1185 \mathrm{~J}\) of heat. The piston moves down and the surroundings do \(623 \mathrm{~J}\) of work on the system. What is \(\Delta E ?\)
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