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Chapter 5: Problem 38

The decomposition of zinc carbonate, \(\mathrm{ZnCO}_{3}(s)\), into zinc oxide, \(\mathrm{ZnO}(\mathrm{s})\), and \(\mathrm{CO}_{2}(\mathrm{~g})\) at constant pressure requires the addition of \(71.5 \mathrm{~kJ}\) of heat per mole of \(\mathrm{ZnCO}_{3}\). (a) Write a balanced thermochemical equation for the reaction. (b) Draw an enthalpy diagram for the reaction.

Short Answer

Expert verified
(a) The balanced thermochemical equation for the decomposition of zinc carbonate is: \(ZnCO_3 (s) \rightarrow ZnO (s) + CO_2 (g);\; \Delta H = 71.5\; kJ/mol\) (b) The enthalpy diagram for this endothermic reaction shows an increase in enthalpy as the reactant (zinc carbonate) is converted into the products (zinc oxide and carbon dioxide). The enthalpy change is 71.5 kJ/mol.

Step by step solution

01

Write the unbalanced equation

Write down the reactants and products in an unbalanced form: \(ZnCO_3 (s) \rightarrow ZnO (s) + CO_2 (g)\)
02

Balance the equation

Since the given reaction is already balanced, it can be rewritten as: \(ZnCO_3 (s) \rightarrow ZnO (s) + CO_2 (g)\)
03

Add the enthalpy change

Include the given enthalpy change (71.5 kJ) to the balanced equation: \(ZnCO_3 (s) \rightarrow ZnO (s) + CO_2 (g)\; \Delta H = 71.5\; kJ/mol\) The balanced thermochemical equation is: \(ZnCO_3 (s) \rightarrow ZnO (s) + CO_2 (g);\; \Delta H = 71.5\; kJ/mol\) (b) Enthalpy Diagram:
04

Label the reactants and products

On the vertical axis, label the enthalpy (H) and on the horizontal axis, label the reaction progress. R should be the reactant side (zinc carbonate) and P should be the product side (zinc oxide and carbon dioxide).
05

Draw the energy levels

Draw a horizontal line to represent the energy level of the reactants (zinc carbonate). Draw another horizontal line above the reactant line representing the energy level of the products (zinc oxide and carbon dioxide).
06

Add the enthalpy change

Draw an arrow between the reactant and product energy levels, with the arrow pointing upward since the reaction is endothermic (enthalpy change is positive). Label this arrow with the enthalpy change of 71.5 kJ/mol.
07

Complete the diagram

Add labels to the energy levels indicating the reactants (zinc carbonate) and the products (zinc oxide and carbon dioxide). The diagram will display an endothermic reaction with an enthalpy change of 71.5 kJ/mol.

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Key Concepts

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

Understanding Enthalpy Change
In chemical reactions, enthalpy change is the amount of heat absorbed or released. For the decomposition of zinc carbonate into zinc oxide and carbon dioxide, the enthalpy change is 71.5 kJ/mol. This means 71.5 kJ of heat is required to decompose one mole of ZnCO3. Enthalpy change is denoted by \( \Delta H \).

If \( \Delta H \) is positive, the reaction absorbs heat, making it endothermic. If \( \Delta H \) is negative, the reaction releases heat, indicating an exothermic process. In this case, since \( \Delta H = 71.5 \) kJ/mol, the reaction absorbs heat from the surroundings.
Exploring Endothermic Reactions
Endothermic reactions are processes that absorb heat energy from their surroundings. In the decomposition of zinc carbonate, since \( \Delta H = 71.5 \) kJ/mol is positive, it's endothermic.

This means the reaction requires energy input to proceed. Such reactions often feel cold to the touch because they take heat from the environment. A simple way to spot endothermic reactions is by observing negative temperature changes in the surroundings post-reaction.
Balancing Chemical Equations
A balanced chemical equation ensures that the number of atoms for each element is the same on both sides of the equation. For the decomposition reaction given in the problem, the equation is:
  • Reactants: \( ZnCO_3 (s) \)
  • Products: \( ZnO (s) + CO_2 (g) \)
No need to adjust coefficients here, as it's already balanced. Correctly balanced equations are crucial as they conform to the law of conservation of mass, meaning matter is not created or destroyed in chemical reactions. This helps in precise calculations of reactants and products involved.

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Most popular questions from this chapter

(a) Why are tables of standard enthalpies of formation so useful? (b) What is the value of the standard enthalpy of formation of an element in its most stable form? (c) Write the chemical equation for the reaction whose enthalpy change is the standard enthalpy of formation of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{~s}), \Delta \mathrm{H}_{f}^{\circ}\left[\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right]\).

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