You measure Ksp for borax and find it to be 6.2 x 10-3 at 22.8°C. If the standard enthalpy change, ΔH°, for the dissolution of borax is 58 kJ, calculate the standard entropy change, ΔS°, (in J/K) for the dissolution of borax at this temperature. (Assume ΔH° is independent of temperature.)
Use dG = -RT*lnK and calculate dG.
Then dG = dH - TdS.
To calculate the standard entropy change, ΔS°, for the dissolution of borax at a given temperature, we can use the following equation:
ΔG° = ΔH° - TΔS°
Where:
ΔG° is the standard Gibbs free energy change
ΔH° is the standard enthalpy change
T is the temperature in Kelvin
ΔS° is the standard entropy change
We're given the value of ΔH° = 58 kJ. However, the temperature is given in Celsius, so we need to convert it to Kelvin.
T = 22.8°C + 273.15 = 295.95 K
Now we can rearrange the equation to solve for ΔS°:
ΔS° = (ΔH° - ΔG°) / T
We know that ΔG° = -RT ln(Ksp), where R is the gas constant (8.314 J/(mol·K)) and Ksp is the solubility product constant.
Plugging in the values and solving for ΔS°:
ΔS° = (58 kJ - (-8.314 J/(mol·K) * 295.95 K * ln(6.2 x 10^-3)) / 295.95 K
Now we can calculate ΔS° using these values.