Concept explainers
Interpretation:
The lattice energy of
Concept Introduction:
Lattice energy:
The amount of energy that is necessary for the conversion of one mole of ionic solid to its constituent ions in gaseous phase is called Lattice energy.
Hess’s law:
The enthalpy change for given set of reactants to the given set of products is the same, whether the process takes place in single or sequence of steps. This is called as Hess’s law.
Enthalpy is generally calculated from the standard enthalpy of formation.
With the
To calculate: The lattice energy of
Answer to Problem 10.83QP
The lattice energy of
Explanation of Solution
Lattice energy
Lattice energy =
=
=
Lattice energy of
The lattice energy of
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Chapter 10 Solutions
CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT
- n Section 10.7, two characteristics of enthalpy changes for reactions are listed. What are these characteristics? Explain why these characteristics are true.arrow_forwardThe lattice energy of KF is 794 kJ/mol, and the interionic distance is 269 pm. The Na—F distance in NaF, which has the same structure as KF, is 231 pm. Which of the following values is the closest approximation of the lattice energy of NaF: 682 kJ/mol, 794 kJ/mol, 924 kJ/mol, 1588 kJ/mol, or 3175 kJ/mol? Explain your answer.arrow_forwardYou discover that you cannot carry out a particular reaction for which you would like the enthalpy change. Does this mean that you will be unable to obtain this enthalpy change? Explain.arrow_forward
- Compare your answers from Exercise 72 to the H values calculated from standard enthalpies of formation in Appendix 4. Explain any discrepancies.arrow_forwardWith a platinum catalyst, ammonia will burn in oxygen to give nitric oxide, NO. 4NH3(g)+5O2(g)4NO(g)+6H2O(g);H=906kJ What is the enthalpy change for the following reaction? NO(g)+32H2O(g)NH3(g)+34O2(g)arrow_forward9.99 The chemical reaction BBr3(g)+BCl3(g)BBr2Cl(g)+BCl2Br(g) , has an enthalpy change very close to zero. Using Lewis structures of the molecules, all of which have a central boron atom, provide a molecular-level description of why H for this reaction might be very small.arrow_forward
- Compare your answers from parts a and b of Exercise 69 of Chapter 3 with H values calculated for each reaction using standard enthalpies of formation in Appendix 4. Do enthalpy changes calculated from bond energies give a reasonable estimate of the actual values?arrow_forwardEnthalpy changes for the following reactions can be determined experimentally: N2(g) + 3 H2(g) 2 NH3(g) rH = 91.8 kJ/mol-rxn 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) rH = 906.2 kJ/mol-rxn H2(g) + O2(g) H2O(g) rH = 241.8 kl/mol-rxn Use these values to determine the enthalpy change for the formation of NO(g) from the elements (an enthalpy change that cannot be measured directly because the reaction is reactant-favored). N2(g) + O2(g) NO(g) rH = ?arrow_forwardIs the following reaction the appropriate one to use in determining the enthalpy of formation of methane, CH4(g)? Why or why not? C(g)+4H(g)CH4(g)arrow_forward
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