Endothermic reactions by definition, require heat as a reactant. By drawing in heat from the surrounding, the surrounding will have a lower temperature compared to air temperature, which is why the ice pack feels cold. If it were an exothermic reaction, the pack would release heat, making the pack feel warm. 

The act of dissolving a solute in a solvent is a local increase in entropy, converting a single molecule to multiple ions. The absorption of heat from the surroundings (cool beaker) indicates that this is an endothermic dissolution. We can look at the equation for Gibbs free energy to evaluate the possible answers.

In order to be spontaneous, the reaction must have a negative Gibbs free energy. To accomplish this, a reaction may have a negative enthalpy (exothermic) and positive entropy, however we know that our reaction has a positive enthalpy (endothermic) and positive entropy. A reaction will be spontaneous if it has a positive  and a positive  only when temperature is high.

The question specifies that this reaction is exothermic, and that it has a positive local increase in entropy as it progresses. This means that the change in enthaply is negative and the change in entropy is positive.

For a reaction to be spontaneous, Gibbs free energy must be negative.

If enthalpy is negative and entropy is positive, the temperature is irrelevant and Gibbs free energy will always be negative.

The reaction is spontaneous at any temperature.

The spontaneity of a reaction is determined by the equation for Gibbs free energy. 

Here, H is a positive number and S is negative, meaning that G will always be a positive value. T will be given in Kelvin, and cannot be negative. Reactions with positive Gibbs free energy values are never spontaneous.

A positive value for ΔG (Gibbs free energy) will guarantee a nonspontaneous reaction. When ΔH (enthalpy) is postive and ΔS (entropy) is negative, the change in Gibbs free energy must be positive and, therefore, nonspontaneous.

Because T (temperature) will always have a positive value, a negative entropy and positive enthalpy will always result in a positive Gibbs free energy.

All of the following scenarios would lead to spontaneous reaction, since each scenario would result in a negative Gibbs free energy (-ΔG).

Negative enthalpy, positive entropy:

Positive enthalpy and entropy with high temperature:

Negative enthalpy and entropy with low temperature:

According to the Gibbs free energy equation, a system is at equilibrium when  is equal to 0.

Since , a system is in equilibrium when               . A negative Gibbs free energy means that the reaction will be spontaneous. At equilibrium, the forward reaction rate equals the reverse reaction rate, though the net rate is zero.

The rate law can be determined by comparing changes in reactant concentration to changes in rate. In this reaction, doubling XY doubles the reaction rate; the reaction is first-order for XY. Doubling Z increases the rate by a factor of eight, so the reaction is third-order for Z (2³ = 8).

Knowing that Rate = k[XY] [Z]³, we can solve for k by using data from one of the trials. Plugging in data from trial 1, we can calculate the value of k.

0.16 = k[100][200]³

k = 2 * 10^-10

Let's look at each of the three reactants individually.

Compare trials 1 and 3. Doubling  without changing the other reactants doubles the rate from to ; therefore  has a superscript of 1.

Compare trials 3 and 4. Doubling  without changing the other reactants results in a rate that is four times faster, increasing from to . The rate increases with the square of concentration of , 

Compare trials 1 and 2. Doubling without changing the other reactants has no effect on the rate, which remains at , so  is not part of the final rate equation.

The final rate law will be .

Comparing trials 2 and 3, we see that when P_XY is constant and P_Z doubles, the rate increases by a factor of 8, or 2³. Rate is therefore proportional to [Z]³.

Comparing trials 1 and 2, we see that when P_Z is held constant and P_XY doubles, the rate doubles. Rate is therefore proportional to [XY].

Remember that in these problems, you are looking for the value of n in the function . When doubling the concentration does not affect the reaction rate, n = 0. When doubling the concentration doubles the reaction rate, n = 1. When doubling the concentration quadruples the reaction rate, n = 2.