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What is \(\Delta H_{\text {rxn }}\) for the reaction between \(\mathrm{H}_{2} \mathrm{S}\) and \(\mathrm{O}_{2}\) that yields \(\mathrm{SO}_{2}\) and water, \(2 \mathrm{H}_{2} \mathrm{S}(g)+3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H_{\mathrm{rxn}}=?\) given \(\Delta H_{\mathrm{rxn}}\) for the following reactions? $$\begin{aligned} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(g) & & \Delta H_{\mathrm{rxn}}=-241.8 \mathrm{kJ} \\ \mathrm{SO}_{2}(g)+3 \mathrm{H}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{S}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) & & \Delta H_{\mathrm{rxn}}=34.8 \mathrm{kJ} \end{aligned}$$

Short Answer

Expert verified
Answer: The enthalpy change for the given reaction is \(-518.4\,\mathrm{kJ}\).

Step by step solution

01

Identify the objective

We want to manipulate and combine the two given reactions to obtain the target reaction: \(2 \mathrm{H}_{2} \mathrm{S}(g)+3 \mathrm{O}_{2}(g) \rightarrow 2\mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\).
02

Multiply the first given reaction

Since the target reaction has two moles of \(\mathrm{H}_{2}\mathrm{O}\), multiply the first given reaction and its enthalpy change by 2: $$2(\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2}\mathrm{O}(g)) \Rightarrow 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2}\mathrm{O}(g)$$ $$2(-241.8\,\mathrm{kJ})=-483.6\,\mathrm{kJ}$$
03

Reverse the second given reaction

The target reaction has \(\mathrm{H}_{2}\mathrm{S}\) as a reactant, but in the second given reaction, it is a product. Reverse the second given reaction and negate its enthalpy change: $$\mathrm{H}_{2}\mathrm{S}(g)+2 \mathrm{H}_{2}\mathrm{O}(g) \rightarrow \mathrm{SO}_{2}(g)+3 \mathrm{H}_{2}(g)$$ $$\Delta H_{\text{rxn}} = -34.8\,\mathrm{kJ}$$
04

Combine the modified reactions and their enthalpy changes

Add the modified first reaction and the reversed second reaction to obtain the target reaction, and add their enthalpy changes to calculate the overall enthalpy change: $$\begin{aligned} (2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2}\mathrm{O}(g))&+(-483.6 \,\mathrm{kJ}) \\ (\mathrm{H}_{2}\mathrm{S}(g)+2 \mathrm{H}_{2}\mathrm{O}(g) \rightarrow \mathrm{SO}_{2}(g)+3 \mathrm{H}_{2}(g)) &+(-34.8 \,\mathrm{kJ}) \\ \hline 2 \mathrm{H}_{2} \mathrm{S}(g)+3 \mathrm{O}_{2}(g) \rightarrow 2\mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) &+ (-483.6 -34.8) \,\mathrm{kJ} \end{aligned}$$
05

Calculate the overall enthalpy change

Add the two enthalpy changes from the above step to find the overall enthalpy change for the target reaction: $$-483.6\,\mathrm{kJ} - 34.8\,\mathrm{kJ} = -518.4\,\mathrm{kJ}$$ Therefore, the enthalpy change for the target reaction, \(\Delta H_{\text{rxn}}\), is \(-518.4\,\mathrm{kJ}\).

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Key Concepts

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

Chemical Reactions
In chemistry, understanding chemical reactions is essential. A chemical reaction involves the rearrangement of atoms, where reactants transform into products.
For instance, in the reaction given: \(2 \ \mathrm{H}_{2} \mathrm{S}(g) +3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{2}(g) + 2 \mathrm{H}_{2} \mathrm{O}(g)\), \ \mathrm{H}_{2} \mathrm{S} \ and \ \mathrm{O}_{2} \ are reactants, while \ \mathrm{SO}_{2} \ and \ \mathrm{H}_{2} \mathrm{O} \ are products. Each chemical reaction can be described by a balanced chemical equation, showing all reactants and products with appropriate stoichiometric coefficients. These coefficients indicate the proportional amount of each substance involved.
Balancing enables the conservation of mass, ensuring that the number of each type of atom is the same on both sides of the reaction equation.
For someone analyzing these reactions, it's vital to recognize how reactants change into products and indicate it clearly in a chemical equation.By understanding how atoms and molecules interact in chemical reactions, scientists can predict reaction outcomes and manipulate reactions to obtain desired products.
Hess's Law
One of the foundational principles in thermochemistry is Hess's Law. It states that the total enthalpy change during a chemical reaction is the same, regardless of the number of steps in the reaction.
Therefore, the enthalpy change is considered a state function, depending only on the initial and final energy states, not the process itself. In practical terms, Hess's Law allows one to calculate enthalpy changes for reactions where direct measurement is difficult. By manipulating and combining given reactions with known enthalpy changes, we can determine the enthalpy change for complex reactions.
  • This process involves reversing reactions, possibly multiplying them by coefficients to match target reactions, and summing up the enthalpy changes.
  • When a reaction is reversed, the enthalpy change sign also reverses.
  • If coefficients are used to balance reactions, the enthalpy change is multiplied by the same factor.
By carefully applying these rules, calculating the enthalpy change for the target reaction becomes straightforward. In our exercise above, Hess's Law provided a clear pathway to ascertain the total enthalpy change.
Thermochemistry
Thermochemistry focuses on the study of energy changes during chemical reactions, emphasizing the heat exchange between a system and its surroundings.
These energy changes are captured as enthalpy changes (\(\Delta H\)), providing insights into whether a process is endothermic or exothermic.
  • An endothermic reaction absorbs energy from its surroundings, resulting in a positive \(\Delta H\).
  • Conversely, an exothermic reaction releases energy, indicated by a negative \(\Delta H\).
In the exercise, the reaction \(2 \ \mathrm{H}_{2} \mathrm{S}(g) +3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{2}(g) + 2 \mathrm{H}_{2} \mathrm{O}(g)\) had a calculated \(\Delta H\) of \(-518.4\,\mathrm{kJ}\), categorizing it as exothermic due to the release of energy. Thermochemistry not only helps in understanding energy dynamics in reactions but also aids in designing processes for industry and research purposes where energy efficiency is crucial.

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Most popular questions from this chapter

When measuring the enthalpy of combustion of a very small amount of material, would you prefer to use a calorimeter having a heat capacity that is small or large? Explain your reasoning.

Why is the standard enthalpy of formation of \(\mathrm{CO}(g)\) difficult to measure experimentally?

The heavier hydrocarbons in white gas are hexanes \(\left(\mathrm{C}_{6} \mathrm{H}_{14}\right)\) a. Calculate the fuel value of \(C_{6} H_{14},\) given that \(\Delta H_{\mathrm{comb}}^{\circ}=-4163 \mathrm{kJ} / \mathrm{mol}\) b. How much energy is released during the combustion of \(1.00 \mathrm{kg}\) of \(\mathrm{C}_{6} \mathrm{H}_{14} ?\) c. How many grams of \(\mathrm{C}_{6} \mathrm{H}_{14}\) are needed to heat \(1.00 \mathrm{kg}\) of water from \(25.0^{\circ} \mathrm{C}\) to \(85.0^{\circ} \mathrm{C} ?\) Assume that all the energy released during combustion is used to heat the water. d. Assume white gas is \(25 \%\) C \(_{5}\) hydrocarbons and \(75 \% \mathrm{C}_{6}\) hydrocarbons; how many grams of white gas are needed to heat \(1.00 \mathrm{kg}\) of water from \(25.0^{\circ} \mathrm{C}\) to \(85.0^{\circ} \mathrm{C} ?\)

Which symbol, \(\Delta H_{\text {comb }}\) or \(\Delta H_{\text {fus }},\) refers to a physical change?

An increasing number of vehicles in the United States can run on either gasoline \(\left[\mathrm{C}_{9} \mathrm{H}_{20}(\ell), \Delta H_{\text {comb., } \text { assoline }}=-6160 \mathrm{kJ} / \mathrm{mol}\right]\) or ethanol \(\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(\ell), \Delta H_{\text {comb,ethanol}}=-1367 \mathrm{kJ} / \mathrm{mol}\right] .\) Which fuel has the greater fuel value?

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