23. The integral heats of solution at 18°C for the various solid modifications of calcium chloride in the indicated quantities of water are given by CaCl2(s) + 400 H;0(1) = CaCl,(400 H;0) СаCla - 2 H:0(s) + 398 H:0(1) — СаCl:(400 H,0) СаCli 4 H,0(8) + 396 H,0(1) — СаCl:(400 H,0) CaCl2 · 6 H20(s) + 394 H;0(1) = CaCl2(400 H20) AH, = -17,990 cal AH2 = -10,030 cal AH3 = - 1830 cal AH, = +4560 cal From these data find the heats of the following hydration reactions: (a) CaCla(s) + 2 H;0(1) = CaCl2 · 2 H,0(s) (b) CaClą · 2 H20(s) + 2 H;0(1) = CaCl, · 4 H20(s) (c) CaCl, · 4 H20(s) + 2 H¿0(1) = CaCl2 · 6 H20(s) (d) CaCl2(s) + 6 H20(1) CaCl2 · 6 H20(s)
23. The integral heats of solution at 18°C for the various solid modifications of calcium chloride in the indicated quantities of water are given by CaCl2(s) + 400 H;0(1) = CaCl,(400 H;0) СаCla - 2 H:0(s) + 398 H:0(1) — СаCl:(400 H,0) СаCli 4 H,0(8) + 396 H,0(1) — СаCl:(400 H,0) CaCl2 · 6 H20(s) + 394 H;0(1) = CaCl2(400 H20) AH, = -17,990 cal AH2 = -10,030 cal AH3 = - 1830 cal AH, = +4560 cal From these data find the heats of the following hydration reactions: (a) CaCla(s) + 2 H;0(1) = CaCl2 · 2 H,0(s) (b) CaClą · 2 H20(s) + 2 H;0(1) = CaCl, · 4 H20(s) (c) CaCl, · 4 H20(s) + 2 H¿0(1) = CaCl2 · 6 H20(s) (d) CaCl2(s) + 6 H20(1) CaCl2 · 6 H20(s)
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Transcribed Image Text:23. The integral heats of solution at 18°C for the various solid modifications
of calcium chloride in the indicated quantities of water are given by
CaCl2(s) + 400 H;0(1)
CaCl2 · 2 H,0(s) + 398 H;0(1)
СаCli 4 H,0(8) + 396 H,0(1) — СаCl:(400 H,0)
CaCl2 · 6 H20(s) + 394 H;0(1) = CaCl2(400 H,0)
AH, = -17,990 cal
AH2 = -10,030 cal
AH3 = -1830 cal
AH, = +4560 cal
СаCla(400 H-0)
СаCli:(400 H:0)
%3D
From these data find the heats of the following hydration reactions:
(a) CaCla(s) + 2 H;0(1) = CaCl, · 2 H20(s)
(b) СаCla - 2 H0(8) + 2 H:0(1) - СаCl; - 4 H,0(s)
(c) CaCl2 · 4 H,0(s) + 2 H¿0(1) = CaCl, · 6 H,0(s)
(d) CaCl2(s) + 6 H20(1)
CaCl2 · 6 H,0(s)
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