Question

The solubility of Drug X at 35^oC is 0.5 moles/L. Predict the solubility of this compound at 25^oC. The ΔH_soln of this compound is “-13700” cal/mol and R is1.987cal/mol/deg.

Answer 1

We know that ln(S'/S) = (-ΔH_soln/R) x [(1/T')-(1/T)]

Where

S = solubility at temerature T (35oC) = 0.5 moles/L

S' = solubility at temerature T (25oC) = ?

T = initial temperature = 35oC = 35+273 = 308 K

T' = final temperature = 25 oC = 25+273 = 298 K

ΔH_soln = -13700 Cal/mol

R = gas constant = 1.987 Cal/mol/deg

Plug the values we get

ln(S'/0.5) = (-(-13700)/1.987) x [(1/298)-(1/308)]

              = 0.751

    S' / 0.5 = e^0.751 = 2.12

        S' = 2.12 x 0.5 = 1.06 moles/L

Therefore the required soluybility is 1.06 moles/L