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(a) What is the value of the standard enthalpy of formation of an element in its most stable form? (b) Write the chemical equation for the reaction whose enthalpy change is the standard enthalpy of formation of sucrose (table sugar), \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s), \Delta H_{f}^{\circ}\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)\right]\)

Short Answer

Expert verified
(a) The standard enthalpy of formation of an element in its most stable form is \(0 \, \text{kJ/mol}\). (b) The balanced chemical equation for the standard enthalpy of formation of sucrose is \(12 C(graphite) + 11 H_{2}(g) + 11 O_{2}(g) \rightarrow C_{12}H_{22}O_{11}(s)\).

Step by step solution

01

(a) Standard Enthalpy of Formation Definition

The standard enthalpy of formation, \(\Delta H_{f}^{\circ}\), is the change in enthalpy when one mole of a substance in its standard state is formed from its constituent elements in their standard states. For an element in its most stable form, it is already in its standard state; thus, no change in enthalpy occurs, and its standard enthalpy of formation is zero.
02

(a) Entropy of Formation Value for a Stable Element

Given the definition of standard enthalpy of formation, we conclude that the value of the standard enthalpy of formation of an element in its most stable form is \(0 \, \text{kJ/mol}\).
03

(b) Formation of Sucrose Chemical Equation

We are asked to write a balanced chemical equation representing the enthalpy change for the formation of sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)\). Sucrose is formed from its constituent elements, carbon (C), hydrogen (H), and oxygen (O), in their standard states (solid graphite, diatomic hydrogen gas, and diatomic oxygen gas, respectively). We will begin by writing the unbalanced chemical equation: \(C(graphite) + H_{2}(g) + O_{2}(g) \rightarrow C_{12}H_{22}O_{11}(s)\) Next, we will balance the equation by adjusting the coefficients to ensure that the number of atoms of each element on the reactant side equals the number on the product side.
04

(b) Balanced Chemical Equation for Sucrose Formation

After balancing the equation, we obtain: \(12 C(graphite) + 11 H_{2}(g) + 11 O_{2}(g) \rightarrow C_{12}H_{22}O_{11}(s)\) This balanced chemical equation represents the standard enthalpy of formation of sucrose, \(\Delta H_{f}^{\circ}\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)\right]\).

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Key Concepts

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

Enthalpy Change
Enthalpy change is a crucial concept in thermodynamics, often symbolized by \( \Delta H \). It represents the heat change at constant pressure, which can occur during a chemical reaction. When a reaction takes place, products form from reactants, and energy either gets absorbed or released. If a system absorbs heat, it's an endothermic reaction, and when it releases heat, it's exothermic.
\( \Delta H \) helps us understand this heat exchange. It's determined by the difference between the enthalpy of the products and the enthalpy of the reactants.
  • To calculate, use: \( \Delta H = H_{products} - H_{reactants} \)
  • A negative \( \Delta H \) indicates an exothermic process.
  • A positive \( \Delta H \) indicates an endothermic process.
By studying enthalpy change, we gain insights into the energetic behavior of substances during reactions, making it a foundational aspect of chemistry.
Standard State
In chemistry, the standard state of a substance is its typical phase and form at a standard set of conditions. These conditions include a pressure of 1 bar (or 1 atm, approximately), and the temperature is usually considered to be 298.15 K (25°C), unless stated otherwise.
This uniform set of conditions allows chemists to compare different properties like enthalpy or entropy across different substances.
  • For gases, the standard state is pure gas at 1 bar.
  • For liquids and solids, it's the pure substance in its most stable form at 1 bar and 25°C.
  • For solutions, the standard state is the solute dissolved at 1 molar concentration.
Understanding the standard state is crucial because it provides a reference point for measuring and reporting various thermodynamic properties, such as the standard enthalpy of formation.
Chemical Equation
A chemical equation represents a chemical reaction where reactants transform into products. It uses chemical formulas to show the substances involved. Each chemical symbol identifies specific elements, and stoichiometric coefficients help balance the equation to ensure mass conservation.
Balancing a chemical equation involves making sure that there are equal numbers of each type of atom on both sides of the equation, which complies with the law of conservation of mass.
  • Start by writing the unbalanced equation.
  • Identify the number of atoms of each element involved.
  • Add coefficients to balance the number of atoms on both sides.
For example, in the formation of sucrose, the balanced chemical equation is:\[12 \ C(graphite) + 11 \ H_{2}(g) + 11 \ O_{2}(g) \rightarrow C_{12}H_{22}O_{11}(s)\]This equation indicates that 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen molecules produce one molecule of sucrose.
Formation of Sucrose
The formation of sucrose involves combining its constituent elements: carbon, hydrogen, and oxygen. Each of these elements exists in a specific standard state. For carbon, this state is as solid graphite. Hydrogen and oxygen are both diatomic gases, represented as \(H_2\) and \(O_2\), respectively.
To form one mole of sucrose \(C_{12}H_{22}O_{11}\), these elements react under standard conditions as described in the balanced chemical equation:
\[\ 12 \ C (graphite) + 11 \ H_{2} (g) + 11 \ O_{2} (g) \rightarrow C_{12} H_{22} O_{11}(s)\]
  • It's essential to start with these elements in their standard states.
  • This equation reflects the change in enthalpy known as the standard enthalpy of formation for sucrose.
  • Understanding formation reactions helps in studying how substances are built from their most fundamental components, providing insight into chemical relationships and energy changes involved.
In essence, knowing how sucrose forms marries concepts like stoichiometry, standard state, and enthalpy change into a coherent understanding of its production process.

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Most popular questions from this chapter

Using values from Appendix \(\mathrm{C}\) , calculate the standard enthalpy change for each of the following reactions: $$ \begin{array}{l}{\text { (a) } 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)} \\ {\text { (b) } \mathrm{Mg}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{MgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)} \\ {\text { (c) } \mathrm{N}_{2} \mathrm{O}_{4}(g)+4 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)} \\ {\text { (d) } \mathrm{SiCl}_{4}(l)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{SiO}_{2}(s)+4 \mathrm{HCl}(g)}\end{array} $$

Consider the combustion of liquid methanol, \(\mathrm{CH}_{3} \mathrm{OH}(l) :\) $$\begin{aligned} \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) & \\ \Delta H &=-726.5 \mathrm{kJ} \end{aligned}$$ (a) What is the enthalpy change for the reverse reaction? (b) Balance the forward reaction with whole-number coefficients. What is \(\Delta H\) for the reaction represented by this equation? (c) Which is more likely to be thermodynamically favored, the forward reaction or the reverse reaction? (d) If the reaction were written to produce \(\mathrm{H}_{2} \mathrm{O}(g)\) instead of \(\mathrm{H}_{2} \mathrm{O}(l),\) would you expect the magnitude of \(\Delta H\) to increase, decrease, or stay the same? Explain.

The complete combustion of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l),\) to form \(\mathrm{H}_{2} \mathrm{O}(g)\) and \(\mathrm{CO}_{2}(g)\) at constant pressure releases 1235 \(\mathrm{kJ}\) of heat per mole of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) (a) Write a balanced thermochemical equation for this reaction. (b) Draw an enthalpy diagram for the reaction.

Consider two solutions, the first being 50.0 \(\mathrm{mL}\) of 1.00 \(\mathrm{MCuSO}_{4}\) and the second 50.0 \(\mathrm{mL}\) of 2.00 \(\mathrm{M} \mathrm{KOH}\) . When the two solutions are mixed in a constant-pressure calorimeter, a precipitate forms and the temperature of the mixture rises from 21.5 to \(27.7^{\circ} \mathrm{C}\) (a) Before mixing, how many grams of Cu are present in the solution of \(\mathrm{CuSO}_{4}\) ? (b) Predict the identity of the precipitate in the reaction. (c) Write complete and net ionic equations for the reaction that occurs when the two solutions are mixed. (d) From the calorimetric data, calculate \(\Delta H\) for the reaction that occurs on mixing. Assume that the calorimeter absorbs only a negligible quantity of heat, that the total volume of the solution is 100.0 \(\mathrm{mL}\) , and that the specific heat and density of the solution after mixing are the same as those of pure water.

Consider the following hypothetical reactions: $$\begin{array}{ll}{\mathrm{A} \longrightarrow \mathrm{B}} & {\Delta H=+30 \mathrm{kJ}} \\ {\mathrm{B} \longrightarrow \mathrm{C}} & {\Delta H=+60 \mathrm{kJ}}\end{array}$$ (a) Use Hess's law to calculate the enthalpy change for the reaction \(A \longrightarrow C .\) (b) Construct an enthalpy diagram for substances \(A,\) and C, and show how Hess's law applies.

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