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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermochemistry
Thermochemistry is the branch of chemistry that studies the heat energy associated with chemical reactions and changes of state. It involves understanding how energy, in the form of heat, is absorbed or released during a chemical reaction. In every chemical process, bonds are broken and formed, leading to changes in energy levels.
The key concept in thermochemistry is enthalpy (\( H \)), which is a measure of the total energy of a thermodynamic system. Enthalpy changes (\( \Delta H \)) are used to quantify the heat change during a reaction at constant pressure:
  • If \( \Delta H \) is negative, the reaction is exothermic and releases heat.
  • If \( \Delta H \) is positive, the reaction is endothermic and absorbs heat.
The goal in solving thermochemistry problems is often to find the enthalpy change for a reaction, such as the exercise, where we analyze and rearrange given reactions to reflect real-life energy changes during the formation of products.
Hess's Law
Hess's Law is a fundamental concept in thermochemistry that states the total enthalpy change for a chemical reaction is the same, no matter how it is carried out in steps. This principle relies on the fact that enthalpy is a state function, depending only on the initial and final states of the system.
Hess's Law allows for the calculation of enthalpy changes for reactions that are difficult to measure directly. By combining known reactions, it is possible to arrive at the enthalpy change for a desired reaction. During this process:
  • Reactions can be reversed, but their enthalpy values must change sign.
  • Equations can be scaled by multiplying to align with the desired stoichiometry, adjusting the enthalpy value accordingly.
This is demonstrated in the exercise by rearranging and scaling given reactions to derive the overall enthalpy change for the target reaction.
Chemical Reactions
Chemical reactions involve rearrangements of atoms to form new substances. Understanding these reactions is crucial in chemistry as they explain how substances transform, producing heat exchange and other observable phenomena.
Reactions can be classified based on their energy profiles like exothermic and endothermic reactions:
  • Exothermic reactions: These release energy, often in the form of heat, resulting in a negative enthalpy change (\( \Delta H < 0 \)).
  • Endothermic reactions: These absorb energy, indicated by a positive enthalpy change (\( \Delta H > 0 \)).
Each reaction in the given exercise showed distinct energy characteristics. By combining these reactions strategically, it was possible to identify the overall energy exchanges involved in the new formation of products. Understanding these principles aids in mastering chemical concepts and energy calculations in reactions.

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Most popular questions from this chapter

Construct a concept map to show the interrelationships between path-dependent and pathindependent quantities in thermodynamics.

Under the entry \(\mathrm{H}_{2} \mathrm{SO}_{4},\) a reference source lists many values for the standard enthalpy of formation. For example, for pure \(\mathrm{H}_{2} \mathrm{SO}_{4}(1), \Delta H_{\mathrm{f}}^{\circ}=-814.0 \mathrm{kJ} / \mathrm{mol}\) for a solution with \(1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) per mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) \(-841.8 ;\) with \(10 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-880.5 ;\) with \(50 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) \(-886.8 ;\) with \(100 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-887.7 ;\) with \(500 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) \(-890.5 ;\) with \(1000 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-892.3 ;\) with \(10,000 \mathrm{mol}\) \(\mathrm{H}_{2} \mathrm{O},-900.8 ;\) and with \(100,000 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-907.3\) (a) Explain why these values are not all the same. (b) The value of \(\Delta H_{\mathrm{f}}^{\circ}\left[\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\right]\) in an infinitely dilute solution is \(-909.3 \mathrm{kJ} / \mathrm{mol} .\) What data from this chapter can you cite to confirm this value? Explain. (c) If \(500.0 \mathrm{mL}\) of \(1.00 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) is prepared from pure \(\mathrm{H}_{2} \mathrm{SO}_{4}(1),\) what is the approximate change in temperature that should be observed? Assume that the \(\mathrm{H}_{2} \mathrm{SO}_{4}(1)\) and \(\mathrm{H}_{2} \mathrm{O}(1)\) are at the same temperature initially and that the specific heat of the \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) is about \(4.2 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1}\).

Some of the butane, \(\mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g}),\) in a \(200.0 \mathrm{L}\) cylinder at \(26.0^{\circ} \mathrm{C}\) is withdrawn and burned at a constant pressure in an excess of air. As a result, the pressure of the gas in the cylinder falls from 2.35 atm to 1.10 atm. The liberated heat is used to raise the temperature of 132.5 L of water in a heater from 26.0 to 62.2 ^ C. Assume that the combustion products are \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) exclusively, and determine the efficiency of the water heater. (That is, what percent of the heat of combustion was absorbed by the water?)

Several factors are involved in determining the cooking times required for foods in a microwave oven. One of these factors is specific heat. Determine the approximate time required to warm \(250 \mathrm{mL}\) of chicken broth from \(4^{\circ} \mathrm{C}\) (a typical refrigerator temperature) to \(50^{\circ} \mathrm{C}\) in a \(700 \mathrm{W}\) microwave oven. Assume that the density of chicken broth is about \(1 \mathrm{g} / \mathrm{mL}\) and that its specific heat is approximately \(4.2 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1}\).

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?

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