Question

6- The true or accepted value for the heat of solution (ΔH soln ) for KNO3 is reported as 34.89 kj/mol. Continuing from the previous problem ( slope = -2696 ) , calculate the accuracy of your result in terms of relative percent error.

The question above is related to this question :

5- assume that the linear fit from the graph above yields a slope of -2696. Calculate the heat of solution .

ln K_sp = -ΔH/R.(1/T) +C where C is a constant

The given equation in

log₁₀K_sp = slope.(1/T) + constant

Note that we need to multiply the first equation by 2.303 to obtain the second one.

From the two equations, we obtain,

-ΔH/R = slope

The value of the slope is given as (-2696). Hence,

-ΔH/R = (-2696)

or, ΔH/R = 2696

or, ΔH = 2696.R

The value of the gas constant is 8.314 J/mol.K

Hence,

ΔH = 2696.(8.314 J/mol) (heat of solution doesn’t contain any temperature related term, I.e, no K^-1).

Therefore, heat of solution

ΔH = 22.414 kJ/mol.

Answer 1

ln Ksp = -ΔH/R.(1/T) +C where C is a constant

The given equation in

log10Ksp = slope.(1/T) + constant

Note that we need to multiply the first equation by 2.303 to obtain the second one.

From the two equations, we obtain,

-ΔH/R = slope

The value of the slope is given as (-2696). Hence,

-ΔH/R = (-2696)

or, ΔH/R = 2696

or, ΔH = 2696.R

The value of the gas constant is 8.314 J/mol.K

Hence,

ΔH = 2696.(8.314 J/mol) (heat of solution doesn’t contain any temperature related term, I.e, no K-1).

Therefore, heat of solution

ΔH = 22.414 kJ/mol.

The relative percent error is therefore 35,8%