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Given the following reactions: \(\mathrm{S}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{SO}_{2}(\mathrm{~g}) \quad \Delta \mathrm{H}=-71.0 \mathrm{Kcal}\) \(\mathrm{SO}_{2}(\mathrm{~g})+(1 / 2) \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{SO}_{3}(\mathrm{~g}) \quad \Delta \mathrm{H}=-23.5 \mathrm{Kcal}\) calculation \(\Delta \mathrm{H}\) for the reaction: \(\mathrm{S}(\mathrm{s})+1(1 / 2) \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{SO}_{3}(\mathrm{~g})\)

Short Answer

Expert verified
The enthalpy change (ΔH) for the reaction \( S(s) + \frac{3}{2} O_{2}(g) \rightarrow SO_{3}(g) \) is -94.5 Kcal.

Step by step solution

01

Write down the given reactions and their ΔH values

\[ \begin{array}{l l l} \text{Reaction 1:} & S(s) + O_{2}(g) \rightarrow SO_{2}(g) & \Delta H_{1} = -71.0 \: \text{Kcal} \\ \text{Reaction 2:} & SO_{2}(g) + \frac{1}{2} O_{2}(g) \rightarrow SO_{3}(g) & \Delta H_{2} = -23.5 \: \text{Kcal} \end{array} \]
02

Add reactions to get the desired reaction

We want to obtain the reaction: \( S(s) + \frac{3}{2} O_{2}(g) \rightarrow SO_{3}(g) \) To get this, we simply add Reaction 1 and Reaction 2: \( S(s) + O_{2}(g) + SO_{2}(g) + \frac{1}{2} O_{2}(g) \rightarrow SO_{2}(g) + SO_{3}(g) \) Cancel out the common terms on both sides of the arrow: \( S(s) + \frac{3}{2} O_{2}(g) \rightarrow SO_{3}(g) \) Now the sum of reactions equals the desired reaction.
03

Calculate ΔH for the desired reaction

Since we added Reaction 1 and Reaction 2 to get the desired reaction, we can also add their ΔH values to find ΔH for the desired reaction: ΔH = ΔH₁ + ΔH₂ = -71.0 Kcal + (-23.5 Kcal) = -94.5 Kcal Therefore, the enthalpy change (ΔH) for the reaction \( S(s) + \frac{3}{2} O_{2}(g) \rightarrow SO_{3}(g) \) is -94.5 Kcal.

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Key Concepts

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

enthalpy change
Enthalpy change, denoted as ΔH, is a crucial concept in chemistry, especially when analyzing chemical reactions. It measures the heat absorbed or released during a reaction at constant pressure. This change in enthalpy can provide insights into whether a reaction is endothermic or exothermic.
  • If ΔH is negative, the reaction releases heat, making it exothermic.
  • If ΔH is positive, the reaction absorbs heat, making it endothermic.
In the context of Hess's Law, the enthalpy change helps assess the energy transformation as reactions are added or reversed. When you sum the ΔH values of given reactions, you can determine the overall enthalpy change for a desired reaction. This is because enthalpy is a state function, meaning it's dependent only on the initial and final states, not the path taken to get there.
Understanding and calculating enthalpy changes allows chemists to predict the feasibility and the heat effect of reactions which is critical for both theoretical studies and practical applications.
chemical reactions
Chemical reactions involve the transformation of one or more substances into new substances. They occur when bonds between atoms in molecules are broken and new bonds are formed, resulting in the creation of different molecules. Chemical reactions are characterized by changes in energy, and often associated with changes in properties, such as color, temperature, or phase.
  • Reactants are the starting substances that undergo change.
  • Products are the new substances formed as a result of the reaction.
The law of conservation of mass dictates that atoms are neither created nor destroyed in a chemical reaction. Therefore, the number of atoms in the reactants is equal to the number of atoms in the products.
In the given exercise, the reaction goes through intermediate steps involving sulfur and oxygen to result in sulfur trioxide (SO₃) from sulfur (S) and oxygen gas (O₂), showcasing how complex reactions can be broken down into simpler reactions.
thermochemistry
Thermochemistry is a branch of chemistry focused on the study of thermal energy changes associated with chemical reactions and physical transformations. It involves the measurement and interpretation of data on energy changes, primarily in the form of heat, accompanying chemical processes.
Thermochemistry abides by the principles of thermodynamics, providing insights into energy transfer, system equilibrium, and the spontaneity of reactions.
  • Exothermic reactions contribute heat to the surroundings.
  • Endothermic reactions absorb heat from the surroundings.
In the study of thermochemistry, Hess's Law becomes a powerful tool, allowing chemists to deduce unknown enthalpy changes using known data from other reactions. It simplifies the process of calculating heat changes by constructing a Hess's Law cycle with several steps leading to an overall reaction, as demonstrated in the step-by-step solution of the given exercise.
This aspect of chemistry is vital in industrial applications and helps in understanding natural processes such as energy transfer within the Earth's atmosphere.

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Most popular questions from this chapter

What is the heat capacity of uranium? Its atomic heat is 26 \(\mathrm{J} /\) mole and its heat capacity is \(238 \mathrm{~g} / \mathrm{mole}\).

At constant volume, the heat capacity of gas differs from the heat capacity at constant pressure. Why?

You have the reaction \(\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\), For both states, 1 mole of water is at \(100^{\circ} \mathrm{C}\) and 1 atm pressure. The volume of 1 mole of water \(=18 \mathrm{ml}, \Delta \mathrm{H}=9710 \mathrm{cal} /\) mole and there are \(24.2\) cal/liter-atm. Calculate the work done in this conversion and the value of \(\Delta \mathrm{E}\).

In the reaction, \(\mathrm{CaCO}_{3}(\mathrm{~s}) \rightarrow \mathrm{CaCO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\) at \(950^{\circ} \mathrm{C}\) and \(\mathrm{CO}_{2}\) pressure of \(1 \mathrm{~atm}\), the \(\Delta \mathrm{H}\) is found to be 176 \(\mathrm{kJ} /\) mole. Assuming that the volume of the solid phase changes little by comparison with the volume of gas generated, calculate the \(\Delta \mathrm{E}\) for this reaction.

The following reaction using hydrogen and oxygen is carried out in a bomb calorimeter: \(2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\ell)\). The following data are recorded: Weight of water in calorimeter = \(2.650 \mathrm{~kg} .\) Initial temperature of water \(=24.442^{\circ} \mathrm{C} .\), Final temperature of water after reaction \(=25.635^{\circ} \mathrm{C}_{.}\), Specific heat of reaction vessel is \(0.200 \mathrm{Kcal}^{1}-\mathrm{kg}\), the weight of the calorimeter is \(1.060 \mathrm{~kg}\), and the specific heat of water is \(1.00\) \(\mathrm{Kcal}^{\circ}-\mathrm{kg}\), calculate the heat of reaction. Assuming that \(0.050\) mole of water was formed in this experiment, calculate the heat of reaction per mole of liquid water formed. Neglect the specific heat of the thermometer and stirrer.

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