The salt cesium perchlorate dissolves in water according to the reaction: CsClO4(s) Cs+(aq) + ClO4-(aq) (a) Calculate the standard enthalpy?

2019-03-20 12:03 am

更新1:

Cesium perchlorate dissolves in water according to reaction CsClO4(s) Cs+(aq) + ClO4-(aq) a.Calculate standard enthalpy change ΔH° for reaction, using following data: CsClO4(s) = -443.1 kJ mol-1 Cs+(aq) = -258.3 kJ mol-1 ClO4-(aq) = -129.3 kJ mol-1 _____ kJ b. Calculate temperature reached by solution formed when 46.0 g of CsClO4 is dissolved in 0.196L of water at 26.0°C. Approximate heat capacity of the solution by heat capacity of 196 g of pure water ignoring the mass of the salt. _____ °C

回答 (2)

胡雪8°

2019-03-20 12:39 am

✔ 最佳答案

CsClO₄(s) ⇌ Cs⁺(aq) + ClO₄⁻(aq) …… ΔH°

ΔH° = ΔHf°[Cs⁺(aq)] + ΔHf°[ClO₄⁻(aq)] - ΔHf° [HClO₄]
= (-258.3) + (-129.3) - (443.1) kJ
= +55.5 kJ

Molar mass of CsClO₄ = (132.9 +35.5 + 16.0×4) g/mol = 232.4 g/mol
Moles of CsClO₄ dissolved = (46.0 g) / (232.4 g/mol) = 0.1979 mol

Heat gained in dissolving CsClO₄ = Heat lost by the water
(0.1979 mol) × (55.5 × 1000 J/mol) = (196 g) × (4.184 J/g°C) × (26.0 - T)°C
26.0 - T = 13.4
T = 12.6
Temperature reached by solution formed = 12.6°C

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Roger the Mole

2019-03-20 12:53 am

a.
products - reactants:
(-258.3 kJ/mol) + (-129.3 kJ/mol) - (-443.1 kJ/mol) = +55.5 kJ

b.
(46.0 g CsClO4) / (232.3561 g CsClO4/mol) x (55500 J/mol CsClO4) / (4.184 J/g·°C) / (196 g) =
13.4°C change

26.0°C - 13.4°C = 12.6°C

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