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Study the following thermochemical data: \(\mathrm{S}+\mathrm{O}_{2} \rightarrow \mathrm{SO}_{2} ; \quad \Delta H=-298.2 \mathrm{~kJ}\) \(\mathrm{SO}_{2}+1 / 2 \mathrm{O}_{2} \rightarrow \mathrm{SO}_{3} ; \quad \Delta H=-98.2 \mathrm{~kJ}\) \(\mathrm{SO}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4} ; \quad \Delta H=-130.2 \mathrm{~kJ}\) \(\mathrm{H}_{2}+1 / 2 \mathrm{O}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O} ; \quad \Delta H=-287.3 \mathrm{~kJ}\) The enthalpy of formation of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) at \(298 \mathrm{~K}\) will be (a) \(-433.7 \mathrm{k} \mathrm{J}\) (b) \(-650.3 \mathrm{~kJ}\) (c) \(+320.5 \mathrm{~kJ}\) (d) \(-813.9 \mathrm{~kJ}\)

Short Answer

Expert verified
-813.9 kJ

Step by step solution

01

Understand the Concept of Enthalpy of Formation

The enthalpy of formation is defined as the change in enthalpy for a reaction that forms one mole of a compound from its elements in their standard states. The standard state is the most stable form of the element at 1 atm pressure and the temperature of interest (usually 298 K).
02

Combine Given Reactions to Form the Target Compound

We need to combine the given reactions such that we get the formation reaction of \( \mathrm{H}_{2}\mathrm{SO}_{4} \) from its elements in their standard states. The standard states are \( \mathrm{S} \) (solid), \( \mathrm{O}_{2} \) (gas), and \( \mathrm{H}_{2} \) (gas).
03

Calculate the Total Enthalpy Change

Combine the given thermochemical equations so that the result matches the formation equation of \( \mathrm{H}_{2}\mathrm{SO}_{4} \). This involves using the given enthalpy changes and ensuring the stoichiometry of the resulting equation is correct.
04

Determine the Enthalpy of Formation of \( \mathrm{H}_{2}\mathrm{SO}_{4} \)

The enthalpy of formation for \( \mathrm{H}_{2}\mathrm{SO}_{4} \) from its elements in their standard states can be found by summing the \( \Delta H \) values of the combined reactions. Ensure all reactions are in the correct stoichiometric ratios.

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Key Concepts

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

Thermochemical Data
Thermochemical data is essential for understanding the energy changes that occur during chemical reactions. It includes values such as enthalpies of reaction, formation, and combustion. These values are determined experimentally and tabulated in reference materials which chemists and chemical engineers frequently consult.

For instance, the enthalpy of reaction, represented as \( \Delta H \), tells us how much heat is released or absorbed when a reaction takes place under constant pressure. A negative value for \( \Delta H \) indicates that the reaction is exothermic (releases heat), while a positive value means it's endothermic (absorbs heat).

Understanding thermochemical data is critical, especially when predicting the energy requirements of industrial processes or the heat released in combustion reactions that power engines and turbines. It's a cornerstone of thermodynamics in chemistry – the study of energy and its transformations.
Standard States in Chemistry
In chemistry, the standard state of a substance is a reference point used to define its enthalpy, entropy, and free energy at a specified temperature (usually 298 K) and pressure (1 atm). The standard state is typically the most stable physical form of the substance at 1 atm and the given temperature.

For example, for the element oxygen, the standard state is O\(_2\) gas at 298 K and 1 atmospheric pressure. For sulfur, it's the solid form. These standard states are used as references when calculating the standard enthalpy of formation, which is the enthalpy change when one mole of a compound is formed from its elements in their standard states.

Using standard states allows chemists to compare the thermodynamic properties of substances under consistent conditions, which is crucial for enthalpy calculations and other thermodynamic analysis.
Enthalpy Change Calculation
The calculation of enthalpy change is a fundamental aspect of chemical thermodynamics. When performing such calculations, it's crucial to carefully consider the stoichiometry of the reactions involved. For the enthalpy of formation, the calculation will be the sum of enthalpy changes for all the steps that lead from the elements in their standard states to the compound in question.

To calculate the overall enthalpy change, we must ensure that both sides of the reaction equation are balanced, meaning that the same number of atoms of each element are present in the reactants and the products. If a reaction needs to be multiplied to balance another, its enthalpy change must also be multiplied accordingly.

For complex formation reactions, this often involves combining multiple steps, each with its respective enthalpy change, to arrive at the overall enthalpy change for the desired reaction. The precision of these calculations is critical, as they inform the feasibility and energy efficiency of chemical processes.

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Most popular questions from this chapter

For a specific work, on an average a person requires \(5616 \mathrm{~kJ}\) of energy. How many kilograms of glucose must be consumed if all the required energy has to be derived from glucose only? \(\Delta H\) for combustion of glucose is \(-2808 \mathrm{~kJ} \mathrm{~mol}^{-1}\). (a) \(0.720 \mathrm{~kg}\) (b) \(0.36 \mathrm{~kg}\) (c) \(0.18 \mathrm{~kg}\) (d) \(1.0 \mathrm{~kg}\)

Given enthalpy of formation of \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{CaO}(\mathrm{s})\) are \(-94.0 \mathrm{~kJ}\) and \(-152 \mathrm{~kJ}\), respectively, and the enthalpy of the reaction: \(\mathrm{CaCO}_{3}(\mathrm{~s}) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\) is \(42 \mathrm{~kJ}\). The enthalpy of formation of \(\mathrm{CaCO}_{3}(\mathrm{~s})\) is (a) \(-42 \mathrm{~kJ}\) (b) \(-202 \mathrm{~kJ}\) (c) \(+202 \mathrm{~kJ}\) (d) \(-288 \mathrm{~kJ}\)

The \(\Delta_{f} H^{\circ}\) for \(\mathrm{CO}_{2}(\mathrm{~g}), \mathrm{CO}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) are \(-393.5,-110.5\) and \(-241.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\), respectively. The standard enthalpy change (in \(\mathrm{kJ}\) ) for the reaction: \(\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) is (a) \(524.1\) (b) \(41.2\) (c) \(-262.5\) (d) \(-41.2\)

Tungsten carbide is very hard and is used to make cutting tools and rock drills. What is the enthalpy of formation (in \(\mathrm{kJ} / \mathrm{mol}\) ) of tungsten carbide? The enthalpy change for this reaction is difficult of measure directly, because the reaction occurs at \(1400^{\circ} \mathrm{C}\). However, the enthalpies of combustion of the elements and of tungsten carbide can be measured easily. \(2 \mathrm{~W}(\mathrm{~s})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{WO}_{3}(\mathrm{~s}) ; \Delta H\) \(=-1680.6 \mathrm{~kJ}\) \(\mathrm{C}(\) graphite \()+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}) ; \quad \Delta H\) \(=-393.5 \mathrm{~kJ}\) \(2 \mathrm{WC}(\mathrm{s})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{WO}_{3}(\mathrm{~s})+2 \mathrm{CO}_{2}(\mathrm{~g})\) \(\Delta H=-2391.6 \mathrm{~kJ}\) (a) \(-38.0\) (b) \(-76.0\) (c) \(-19.0\) (d) \(-1233.8\)

The intermediate \(\mathrm{SiH}_{2}\) is formed in the thermal decomposition of silicon hydrides. Calculate \(\Delta H_{\mathrm{f}}^{\circ}\) of \(\mathrm{SiH}_{2}\) from the following reactions: \(\mathrm{Si}_{2} \mathrm{H}_{6}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{SiH}_{4}(\mathrm{~g})\) \(\Delta H^{\circ}=-11.7 \mathrm{~kJ} / \mathrm{mol}\) \(\mathrm{SiH}_{4}(\mathrm{~g}) \rightarrow \mathrm{SiH}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\) \(\Delta H^{\circ}=+239.7 \mathrm{~kJ} / \mathrm{mol}\) \(\Delta H_{\mathrm{f}}^{\circ}, \mathrm{Si}_{2} \mathrm{H}_{6}(\mathrm{~g})=+80.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (a) \(353 \mathrm{~kJ} / \mathrm{mol}\) (b) \(321 \mathrm{~kJ} / \mathrm{mol}\) (c) \(198 \mathrm{~kJ} / \mathrm{mol}\) (d) \(274 \mathrm{~kJ} / \mathrm{mol}\)

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