This is a Chem lab for Thermodynamics of Borax Solubility Delta G = Delta H - (T

Question

This is a Chem lab for Thermodynamics of Borax Solubility

Delta G = Delta H - (T) Delta S

R = 8.31447215

ln(K) = - (Delta H/ R) * (1/T) + (Delta S/ R)

Group Temperature C) Ksp T (K) 0.005 0.009 11.2 0.004 9.7 21.0 0.011 21.7 0.008 23.5 0.015 40.0 0.156 50.0 1.052 0.819 50.0 Group perature C) Ksp 0.005 2 10.5 0.006 r 22.1 0.011 4 40.0 0.156 T 0.935 50.0 Ln(Ksp) 273.7 0.00365364 -5.3144467 284.2 0.00351865 4.7105307 282.7 0.00353732 5.5375903 294.0 0.00340136 4.55638 294.7 0.00339328 4.8408925 296.5 0.00337268 4.2267338 313.0 0.00319489 1.856618 323.0 0.00309598 0.0504079 & 323.0 0.00309598 0.2002819 0.0032 0.0034 0.0033 1/Temperature (K) 0.0036 R2 0.8851 0.0037

Explanation / Answer

The linear equation is

ln Ksp = (-H/R)*1/T + S/R ….(1)

The regression equation (equation of the plot) is obtained as

y = -10449x + 31.669 …..(2)

Comparing the two equations above, we must have

(-H/R) = -10449 and S/R = 31.669

Since ln Ksp in (1) is dimensionless, we must have the right side of (1) as dimensionless. Since 1/T has the unit of K-1, therefore, the slope will have unit K.

Therefore,

H/R = 10449 K

==> H = (10449 K)*R

==> H = (10449 K)*(8.31447215 J/mol.K) = 86877.92 J/mol = (86877.92 J/mol)*(1 kJ/1000 J) = 86.8779 kJ/mol 86.878 kJ/mol (ans).

Again,

S/R = 31.669

===> S = (31.669)*R

===> S = (31.669)*(8.31447215 J/mol.K) = 263.3110 J/mol.K 263.311 J/mol.K (ans).

Assume H and S to be independent of temperature and hence we can write

G = H – T*S

===> G = (86.878 kJ/mol) – (298 K)*(263.311 J/mol.K) = (86.878 kJ/mol) – (78466.678 J/mol) = (86.878 kJ/mol) – (78466.678 J/mol)*(1 kJ/1000 J) = (86.878 kJ/mol) – (78.467 kJ/mol) 12.411 kJ/mol (ans).

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