Gases tend to have greater entropy than liquids (or solutions), that tend to have greater entropy than solids. Any change that leads to a greater number of gaseous moles tends to lead to an increase in the entropy of the system.

Thus the correct answer here is 2NI₃ (s) → N₂ (g) + 3I₂ (g), in which 2 moles of solid become 4 moles of gas.

Standard enthalpy of hydration is the enthalpy change when one mole of aqueous solution is formed from ions in the gaseous state.

Ions with smaller size and higher charge will have greater enthalpy of hydration values. Therefore in group 1; Li⁺ > Na⁺ > K⁺. And group 2 ions (like Be²⁺) will have significantly higher enthalpy values.

The correct answer (increasingly exothermic) is therefore K⁺, Na⁺, Li⁺, Be²⁺

This phase/state change involves the making of bonds, so the process in exothermic which means that the enthalpy change is negative - so enthalpy decreases.

Gases tend to have greater entropy than liquids (or solutions), that tend to have greater entropy than solids. Therefore entropy also decreases for this process.

The correct answer is Enthalpy decreases and entropy decreases

Standard enthalpy of atomisation is a definition that needs to be learned:

Standard enthalpy of atomisation is the enthalpy change when one mole of gaseous atoms is formed from the element in its standard state (at 100kPa and 298K).

Thus ½At_2(s) → At_(g) is the correct answer, since one mole of gaseous atoms are formed (from the element in its standard state).

Lattice enthalpy is a definition that needs to be learned:

Lattice enthalpy is the enthalpy change when one mole of an ionic compound in its standard state (solid) forms gaseous ions.

Thus KCl (s) → K⁺(g) + Cl⁻(g) is the correct answer.

For a reaction to be thermodynamically spontaneous the change in Gibb's free energy (ΔG) must be equal to or less than zero.

ΔG = ΔH−TΔS (this should be learned)

(Note for explanation: |x| means magnitude of x, that is the numerical value of x without a sign. And remember that temperature, T in the expression is given in Kelvin (K) so is always positive.)

For the reaction to be become spontaneous at lower temperature, ΔH and ΔS would both need to be negative; then ΔG will be negative when |TΔS| < |ΔH|; this will occur as temperature decreases, as |TΔS| becomes smaller as temperature becomes smaller.

Therefore the correct answer is ΔH^o is negative ; ΔS^o is negative

Notes on incorrect answers

If ΔH is positive and ΔS is positive then ΔG will be negative when |TΔS| > |ΔH|; this will occur as temperature increases, as |TΔS| becomes greater as temperature becomes greater.

For the reaction to always be spontaneous (at any temperature), ΔH would need to be negative and ΔS would need to be positive; then ΔG will always be negative.

For the reaction to never be spontaneous (at any temperature), ΔH would need to be positive and ΔS would need to be negative; then ΔG will always be positive.

It is worth learning the energy cycle (shown with a generic metal, M, and non-metal, X):

Following the alternative route, ΔH solution is therefore equal to lattice enthalpy plus ΔH hydration for the ions.

In MgCl₂, there is one Mg²⁺ ion and two Cl⁻ ions, so:

ΔH solution (MgCl₂) = +2540 + (−1963) + 2(−359)

Which gives the correct answer 2540 − 1963 + 2(−359)

Lattice enthalpy as shown on the diagram begins at MgF₂ (s) and ends at Mg²⁺ (g) + 2F⁻ (g), therefore we can take the alternative route beginning with going against the −1124 arrow and following around the energy cycle, inverting the signs of the energy changes when going against the arrows:

Thus, ΔH_LE(MgF₂) = −(−1124) + 148 + 2(79) + 738 + 1450 + 2(−328), which is therefore the correct answer.

Lattice enthalpy increases with increasing charge of the ions and with decreasing size of the ions.

Solubility in water does depend on the relative magnitude of the lattice enthalpy and hydration enthalpy, as shown on the diagram below:

Therefore 1 and 3 only is the correct answer.

The reaction is exothermic, so ΔH must be negative.

The reaction involves 1 mole of liquid and 3 moles of gas producing 5 moles of gas as written, so there must be a gain in entropy, since entropy increases gas>liquid>solid. Therefore ΔS must be positive.

For a reaction to be thermodynamically spontaneous the change in Gibb's free energy (ΔG) must be equal to or less than zero.

ΔG = ΔH−TΔS (this should be learned)

ΔH is negative and ΔS is positive, so ΔG will always be negative, since −TΔS will always be negative. Therefore the reaction will always be spontaneous (at any temperature).

ΔH^o is negative ; ΔS^o is positive ; the reaction is spontaneous is therefore the correct answer.

ΔG = ΔH−TΔS (this should be learned)

(Note for explanation: |x| means magnitude of x, that is the numerical value of x without a sign. And remember that temperature, T in the expression is given in Kelvin (K) so is always positive.)

If ΔH and ΔS are both negative; then ΔG will be negative when |TΔS| < |ΔH|; this will occur as temperature decreases, as |TΔS| becomes smaller as temperature becomes smaller.

Therefore the correct answer is ΔG^o is negative at low temperatures

Notes

For a reaction to be thermodynamically spontaneous the change in Gibb's free energy (ΔG) must be equal to or less than zero.

If ΔH is positive and ΔS is positive then ΔG will be negative when |TΔS| > |ΔH|; this will occur as temperature increases, as |TΔS| becomes greater as temperature becomes greater.

If ΔH is negative and ΔS is positive; then ΔG will always be negative.

If ΔH is positive and ΔS is negative; then ΔG will always be positive.

First electron affinity and second electron affinity are definitions that need to be learned:

First electron affinity is the enthalpy change when one electron is gained by each atom in one mole of gaseous atoms.

Second electron affinity is the enthalpy change when one electron is gained by each 1− ion in one mole of gaseous ions.

Thus S⁻_(g) + e⁻ → S²⁻_(g) is the correct answer.