(a)
Interpretation:
Enthalpy of formation of Silver chloride has to be calculated by using Born-Haber cycle.
Concept Introduction:
Born-Haber cycle:
The enthalpy of formation of ionic crystals are calculated by addition of enthalpies of atomization, ionization, affinity and lattice enthalpy, it is given by Born-Haber so it is called as Born-Haber cycle.
(a)
Explanation of Solution
Enthalpy of formation of Silver chloride is calculated as,
Hence, the enthalpy of formation of Silver chloride is
(b)
Interpretation:
Enthalpy of formation of Sodium chloride has to be calculated by using Born-Haber cycle.
Concept Introduction:
Refer part (a).
(b)
Explanation of Solution
Enthalpy of formation of Silver chloride is calculated as,
Hence, the enthalpy of formation of Silver chloride is
(c)
Interpretation:
The common factor for the differences in
Concept Introduction:
Refer part (a).
(c)
Explanation of Solution
In the ionization and sublimation endothermic process of both compounds are,
From the given vales,
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Chapter 5 Solutions
Chemistry: The Molecular Science
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- (ii) The industrial process for producing sulphuric acid, H2SO4, from elemental sulphur has three steps, using the information below; (a) identify the three steps, and describe the nature of each step (in one phrase); (b) write an equation for the overall reaction. (c) using the data given below, calculate the enthalpy change for the overall reaction. S(s) + O2(g) SO-(g) A;H° = -296.8 kJ 2S0:(g) 2SO2(g) + O2(9) A;H° = -198.2 kJ SO((g) + H2O(1) H2SO4() A;H° = -227.7 kJarrow_forward(c) Using Hess Law and based on the equations, find enthalpy changes, AH, for this reaction;B N2H4() + H2(g) → 2NH3(g) AHr =? N2H4() +O2(g) → N2(g) + 2HO (g) AP = -543 kJ/mol 2H2(g) + O2(g) → 2H2O (g) AP =-484 kJ/mol N2(g) + 3H2(g) 2NH:(g) AP = -92.2 kJ/molarrow_forwardThe reaction 2 Ti (s) + 3 I2 (g) –→2 Til3 (s) having AH = -839 kJ/mol is: (a) endothermic and, thus, absorbs heat from the surroundings. (b) endothermic and, thus, releases heat to the surroundings. (c) exothermic and, thus, absorbs heat from the surroundings. (d) exothermic and, thus, releases heat to the surroundings.arrow_forward
- How much energy (at constant temperature and pressure) must be supplied as heat to 10.0 g of chlorine gas (as Cl2) to produce a plasma (a gas of charged particles, in this case ions) composed of Cl− and Cl+ ions? The enthalpy of ionization of Cl(g) is +1257.5 kJ mol−1 and its electron-gain enthalpy is −354.8 kJ mol−1. ________ kJ. 4 sig. number.arrow_forward(b) What is the first law of thermodynamic? Define the Hess law. Given the following data; 2CIF(g) + 0,(g) CI,0(g) + F,O(g) AH = 167.4 kJ AH = 341.4 kJ SH = -43,4 kJ 2CIF,(g) + 20,(g) → CĻ0(g) + 3F,O(g) 2F,(2) + O(8) → 2F,O(g) calculate AH for the reaction CIF(g) + F(8) CIF,(g)arrow_forwardFrom the following reactions, determine the reactions that are exothermic. (i) F(g)+e →F(8) (ii) CaCl, (s) → Ca(s)+Cl, (g) (iii) NaCl(s)→ Na" (g)+Cl" (g) (iv) K(g)+e →K (g) (A) (i) and (iv) are exothermic (B) (ii) and (iii) are exothermic (C) All the reactions are exothermic (D) (i), (ii), and (iii) are exothermicarrow_forward
- a) In the late eighteenth century Priestley prepared ammonia by reacting HNO3(g) with hydrogen gas. The thermodynamic equation for the reaction is HNO3(g) + 4H2(g) → NH3(g) + 3H2O(g) ΔH = –637 kJ Calculate the amount of energy released when one mole of hydrogen gas reacts. Consider this to be a positive value. b) How much energy is released when 20.00 g of NH3(g) is made to react with an excess of steam to form HNO3 and H2 gases? Again, enter this as a positive value. HNO3(g) + 4H2(g) → NH3(g) + 3H2O(g) ΔH = –637 kJarrow_forwardA chemist measures the enthalpy change AH during the following reaction: 2 Al(OH),(s)-Al,0,(s) + 3H,0(1) AH=41. kJ Use this information to complete the table below. Round each of your answers to the nearest kJ/mol. reaction ΔΗ 3AI,0,(s) + 9H,0() 6Al(OH),(s) Al,0,(s) + 3H,0() 2Al (OH), (s) Al(OH),(6) - Al,0,(0) + ,00 1 Al,0,(5) + H,0() O kJ Explanation Check 2021 McGra M Barrow_forwardCalculate Δ Hrxn for Ca(s) + 1/2O2(g) + CO2(g) -> CaCO3(s) Given the following set of reactions: Ca(s) + 1/2O2(G) -> CaO(2) ΔH= -635.1kJ CaCO3(s) -> CaO(s) + CO2(g) ΔH = 178.3kJ a.) 813.41 kJ b.) -813.4 kJ c.) 456.8 kJ d.) -456.8 kJarrow_forward
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