A solution of salt water is prepared by adding 100 g of NaCl(s) initially at 25

Question

A solution of salt water is prepared by adding 100 g of NaCl(s) initially at 25 degree C to 1 L of pure water at 90 degree C. Assuming the final temperature of the solution remains at 90 degree C, calculate delta G, delta H, and AS for this process. ii) Re-calculate the results in part (i) for the case of mixing at 25 degree C. Given: H_2O: delta H_f = -258.8 kJ/mol c_P = 75.327 J/mol K S_0 = 69.95 J/mol K NaCl: delta H_f = -411.1 kJ/mol c_P = 50.72 + 6.67 middot 10^-3 middot T - 2.52 middot 10^-6 middot T^2 J/mol K S_0 = 72.08 J/mol K

Explanation / Answer

i)
Find Hrxn at 25oC,
Hrxn = Hproducts - Hreactants , since they never dissociated Hproducts =0
Hrxn = [ 0- [1mol(-411.1kJ/mol) + 1mol(-258.8) ] = 669.9 kJ
At 90oC QNacl = (100gm / 40gm/mol) * [50.72+6.67*10-3(363.15)-2.52*10-6(363.15)2] * (363.15-298.15)
QNaCl = 8689.6131 J = 8.6896 kJ
At 90oC QH2O = ( (1lit * 965.3 g/lit ) / 18 gm/mol) * 75.327 * (363.15-298.15)
QH2O = 262.5752 kJ
Therefore H = Hrxn +  QNacl + QH2O = 669.9kJ +  8.6896 kJ +  262.5752 kJ = 941.168 kJ
S = nSoproducts – nSoreactants
S = - (1 mol *72.08J/mol K +1mol*( 69.95 J/mol K)) = -142.03 J/k = 0.14203 kJ/K
G =  H -TS = 941.168 kJ - 363.15 K * (0.14203 kJ/K) = 889.589 kJ

ii) mixG = nRT (x1ln x1 + x2 ln x2)  
n1 = (1lit * 1000 g/lit ) / 18 gm/mol) = 55.55 moles ; n2 =  (100gm / 40gm/mol) = 2.5 moles
n = 55.55 + 2.5 = 58 moles ; x1 = 55.55 / 58 = 0.95775 ; x2 = 2.5/58 = 0.043103
mixG = nRT (x1ln x1 + x2 ln x2) = 58 * 8.314 * 298 (0.95775 ln(0.95775)+0.043103 ln(0.043103))
mixG = -25.4157 kJ
mixS = -nR(x1 ln x1 + x2 ln x2) = 0.085287 kJ
mixG =  mixH -TmixS
-25.4157 kJ + 298 K ( 0.085287 kJ) = mixH = -1.74*10-4 kJ

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