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Chapter 7: Problem 72

Determine \(\Delta H^{\circ}\) for this reaction from the data below. \(\mathrm{N}_{2} \mathrm{H}_{4}(1)+2 \mathrm{H}_{2} \mathrm{O}_{2}(1) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(1)\) $$\begin{array}{r} \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{l})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta H^{\circ}=-622.2 \mathrm{kJ} \end{array}$$ $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H^{\circ}=-285.8 \mathrm{kJ}$$ $$\mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}_{2}(1) \quad \Delta H^{\circ}=-187.8 \mathrm{kJ}$$

Short Answer

Expert verified
\(\Delta H^{\circ} = -1341.0\, kJ\) for the final reaction.

Step by step solution

01

Arrange the Reactions

First, ensure that the reactions align with the final reaction. The second reaction needs to be multiplied by 2 so that the product, \(H_{2}O(l)\), aligns with the final reaction: \(2H_{2}(g) + O_{2}(g) \longrightarrow 2H_{2}O(l), \Delta H^{\circ} = -2(285.8\, kJ)\)
02

Cancel Out Unwanted Reactants and Products

Notice that the product of the last reaction, \(H_{2}O_{2}(l)\), is not present in the final reaction. This is also the case for the reactant \(O_{2}(g)\) in the third reaction. So, this means we must subtract the third reaction from the desired one to cancel these out: \(\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{l}) + 2\,\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{l}) - [2 \mathrm{H}_{2}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g})] = \mathrm{N}_{2}(\mathrm{g}) + 4\,\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) - \mathrm{H}_{2} \mathrm{O}_{2}(1)\)
03

Add Up the Enthalpies

After correctly aligning, adding and subtracting the reactions, we need to do the same with their respective enthalpy changes. So, \(\Delta H^{\circ}\) for the final reaction would equal \(-622.2\, kJ - 2(285.8\, kJ) + 187.8\, kJ\).

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Most popular questions from this chapter

The enthalpy of sublimation ( solid \(\rightarrow\) gas) for dry ice (i.e., \(\mathrm{CO}_{2}\) ) is \(\Delta H_{\mathrm{sub}}^{\circ}=571 \mathrm{kJ} / \mathrm{kg}\) at \(-78.5^{\circ} \mathrm{C} .\) If \(125.0 \mathrm{J}\) of heat is transferred to a block of dry ice that is \(-78.5^{\circ} \mathrm{C},\) what volume of \(\mathrm{CO}_{2} \operatorname{gas}(d=1.98 \mathrm{g} / \mathrm{L})\) will be generated?

In your own words, define or explain the following terms or symbols: (a) \(\Delta H ;\) (b) \(P \Delta V ;\) (c) \(\Delta H_{f} ;\) (d) standard state; (e) fossil fuel.

In a student experiment to confirm Hess's law, the reaction $$\mathrm{NH}_{3}(\text { concd aq })+\mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{aq})$$ was carried out in two different ways. First, \(8.00 \mathrm{mL}\) of concentrated \(\mathrm{NH}_{3}(\text { aq })\) was added to \(100.0 \mathrm{mL}\) of 1.00 M HCl in a calorimeter. [The NH \(_{3}(\) aq) was slightly in excess.] The reactants were initially at \(23.8^{\circ} \mathrm{C},\) and the final temperature after neutralization was \(35.8^{\circ} \mathrm{C} .\) In the second experiment, air was bubbled through \(100.0 \mathrm{mL}\) of concentrated \(\mathrm{NH}_{3}(\mathrm{aq})\) sweeping out \(\mathrm{NH}_{3}(\mathrm{g})\) (see sketch). The \(\mathrm{NH}_{3}(\mathrm{g})\) was neutralized in \(100.0 \mathrm{mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\). The temperature of the concentrated \(\mathrm{NH}_{3}(\text { aq })\) fell from 19.3 to \(13.2^{\circ} \mathrm{C} .\) At the same time, the temperature of the 1.00 M HCl rose from 23.8 to 42.9 ^ C as it was neutralized by \(\mathrm{NH}_{3}(\mathrm{g}) .\) Assume that all solutions have densities of \(1.00 \mathrm{g} / \mathrm{mL}\) and specific heats of \(4.18 \mathrm{Jg}^{-1 \circ} \mathrm{C}^{-1}\) (a) Write the two equations and \(\Delta H\) values for the processes occurring in the second experiment. Show that the sum of these two equations is the same as the equation for the reaction in the first experiment. (b) Show that, within the limits of experimental error, \(\Delta H\) for the overall reaction is the same in the two experiments, thereby confirming Hess's law.

A 465 g chunk of iron is removed from an oven and plunged into \(375 \mathrm{g}\) water in an insulated container. The temperature of the water increases from 26 to \(87^{\circ} \mathrm{C}\). If the specific heat of iron is \(0.449 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1},\) what must have been the original temperature of the iron?

Use Hess's law to determine \(\Delta H^{\circ}\) for the reaction $$\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}), \text { given that }$$ $$\begin{array}{l} \text { C(graphite) }+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g}) \\ &\left.\qquad \Delta H^{\circ}=-110.54 \mathrm{k} \mathrm{J}\right] \end{array}$$ $$\begin{aligned} &\text { C(graphite) }+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})\\\ &&\Delta H^{\circ}=-393.51 \mathrm{kJ} \end{aligned}$$
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