Determine the temperature change when 4.00 g of (a) KCl; (b) MgBr2; (c) KNO3; (d

Question

Determine the temperature change when 4.00 g of (a) KCl; (b) MgBr2; (c) KNO3; (d) NaOH is dissolved in 100. g of water. Assume that the specific heat capacity of the solution is 4.18 J/K*gand that the enthalpies of solution in Table 5D.3 are applicable.

TABLE 5D.3 Limiting Enthalpies of Solution, AHsol/(kJ mol 1), at 25 °C Anion Cation fluoride chloride bromide iodide hydroxide carbonate sulfate nitrate lithium +4.9 37.0 48.8 63.3 23.6 18.2 2.7 29.8 sodium +1.9 +3.9 0.6 7.5 44.5 26.7 +20.4 2.4 potassium 17.7 17.2 19.9 +20.3 57.1 30.9 +34.9 23.8 ammonium 1.2 14.8 16.0 +13.7 +25.7 +6.6 silver 22.5 +65.5 +84.4 112.2 41.8 +22.6 +17.8 magnesium 12.6 160.0 185.6 213.2 +2.3 25.3 90.9 91.2 calcium 11.5 81.3 103.1 119.7 16.7 13.1 19.2 18.0 aluminum 27 329 368 385 350

Explanation / Answer

Hsoln = heat capacity of water × weight of water × T

Therefore ,

T=((Hsoln)/mol.wt)×4)/(heat cap water×mass of wat)

1.KCl

Molar mass = 74.56g/mole

Hsoln= 17.2KJ = 17200J

T,T1 -T2 =( (17200/74.56)×4)/(4.18 ×100)

= 2.21 K

The tempersture of soln will reduce for 2.21K or

2.MgBr2

Molar mass = 184.113

Hsoln = -185.6KJ = 185600 J

T, T1 - T2 =(( -185600/184.113)×4)/(4×100)

=-9.64K

The temperature of soln will increase for 9.64K or

3.KNO3

Molar mass=101.103g/mol

Hsoln = - 23.8KJ/mol=-23800J

T , T1 -T2 = ((-23800/101.103)×4))/(4.18×100)

= -2.25K

The temperature of solution will increase for 2.25K or

4.NaOH

Molar mass = 40

Hsoln = -44.5KJ = 44500 J

T = ((-44500/40)×4)/(4.18×100)

= -10.64 K

The temperature of solution will increse for 10.64K or

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