Question

THER 112 The Themodynamics of Porasiun Nirate DingWe Post-Laboratory Questions (Use the spaces provided for the answers and additional paper it necessary) 1. (a) Is the process of KNO, dissolving in water spontaneous at all tomperatures studied? Brielly 3. (a) What assumption is made about the reac- tion at the temperature at which crystais become visible? explain, (b) Is the reaction in (a) one that gives oft heat or requires heat? Briefly explain. (b) If this assumption were not true, how would the results tor AG change? (c) is your value of AS consistent with the ex- pected change in disorder for the reaction in Equation 1? Briefly explain. 2. A few compounds exist whose solubility de- creases as the temperature increases. How would the Values for Δ a, ΔΗ, and Δ.Stor these reactions be dif- ferent from those values observed for the solubility of KNO,? Briefly explain.

Answer 1

Please correct the values for $\Delta G^{\circ}$ , the magnitude is correct but the sign should be negative because, $\Delta G^{\circ}$ = - RTlnK_sp. As R, T and lnK_sp are positive values so, value of $\Delta G^{\circ}$ should have been negative.

Answer 1 a)

Spontaneity of a reaction depends on $\Delta G^{\circ}$ , if this value is positive, reaction is non spontaneous. If it is negative, reaction is spontaneous.

As value of $\Delta G^{\circ}$ is negative at all temperatures, the reaction is spontaneous at all temperatures.

Answer 1 b)

Reaction for which $\Delta H$ is positive is said to be endothermic. If it is negative, then the reaction is exothermic.

As the calculated value of $\Delta H$ is positive, the reaction is endothermic i.e. it requires heat.

Answer 1 c)

Value of $\Delta S$ is positive in all cases i.e. entropy increases with dissolution.

This is as per expectation because from the equation:

KNO₃ (s) + H₂O(excess) --------> K⁺ (aq.) + NO₃^- (aq.)

It is clear that the ions are free to move in aqueous medium i.e. in disordered state but before dissolution, these ions were restricted to lattice point i.e. ordered state. So, dissolution has increased randomness or disorder, which would result in positive value of $\Delta S$.

$\Delta S$ can be calculated as:

$\Delta S$ = S_K+ + S_NO3- - S_KNO3

Answer 2

For those compounds whose solubility decreases with increase in temperature,

value of $\Delta G^{\circ}$ will decrease (becomes less negative) with increase in temperature.

$\Delta H$ will be negative i.e. reaction would be exothermic, it will release heat on dissolution.

$\Delta S$ values would have increased with increase in temperature just like KNO₃ because the ions will get kinetic energy

Answer 3 a)

Assumption made is that the solution becomes saturated with KNO₃ at this temperature and the solid is in equilibrium with the solution.

Answer 3 b)

If the assumption was not true, equilibrium would not have been achieved. So, Ksp value calculated would have been lesser than actual value and hence $\Delta G^{\circ}$ calculated would not have been lesser in magnitude in comparison to actual.