Spontaneity, or the likelihood that a reaction will occur, is determined by Gibbs free energy. The equation for Gibbs free energy is:

$\displaystyle \Delta G=\Delta H-T\Delta S$

$\displaystyle \Delta S$ is the term for entropy and a negative value for Gibbs free energy indiciates a spontaneous reaction. Thus, as temperature increases the effective value of the entropy term increases as well (since $\displaystyle \Delta S$ is positive). Since the enthalpy,$\displaystyle \Delta H$, remains constant, increasing the entropy term will have the total effect of decreasing the Gibbs free energy since entropy is subtracted from enthalpy. Decreasing the Gibbs free energy will result in a more spontaneous reaction.

The spontaneity of a reaction can be determined using the Gibb's free energy equation:

$\displaystyle \Delta G = \Delta H - T\Delta S$

In this formula, $\displaystyle \small \Delta H$ is enthalpy, $\displaystyle \small \Delta S$ is entropy, and $\displaystyle \small T$ is the temperature in Kelvin. In order for a reaction to be spontaneous, Gibb's free energy must be negative. Looking at the equation, we can see that the value for $\displaystyle \small \Delta G$ will ALWAYS be negative if enthalpy is negative and entropy is positive.

$\displaystyle (-)-T(+)=\text{negative}$

These are the requirements for a reaction that is always spontaneous, regardless of temperature,

The first thing we need to do in order to solve this problem is determine at what temperature $\displaystyle \small \Delta G$ is equal to zero. When $\displaystyle \small \Delta G$ is equal to zero, the reaction is at equilibrium and will not go one way or the other. Using the Gibb's free energy equation, we can determine the temperature where the reaction is at equilibrium.

$\displaystyle \Delta G = \Delta H - T\Delta S$

The question tells us the enthalpy and entropy values, allowing us to solve for the equilibrium temperature. Remember to convert the enthalpy from kilojoules to Joules.

$\displaystyle \small \Delta H=28.05kJ$   $\displaystyle \small \Delta S=108.7\frac{J}{K}$

$\displaystyle 0J=29050J-T(108.7\frac{J}{K})$

$\displaystyle T = 258K$

At $\displaystyle \small 258K$, the reaction is at equilibrium. In order to make the reaction spontaneous, do we need to raise or lower from this temperature? Notice how entropy in this case is positive. By increasing temperature, the negative portion of the equation will become larger, and will result in a negative value. This means that the above reaction is spontaneous at temperatures higher than $\displaystyle \small 258K$. You can check you work by solving for the Gibb's free energy at a higher temperature.

$\displaystyle \Delta G = \Delta H - T\Delta S$

$\displaystyle \Delta G=29050J-(300K)(108.7\frac{J}{K})$

$\displaystyle \Delta G=-3560J$

Since the value is negative, the reaction is spontaneous at this higher temperature.

Given the above conditions, we can use the Gibb's free energy equation in order to determine the reaction's spontaneity.

$\displaystyle \Delta G = \Delta H - T\Delta S$

Convert the temperature to Kelvin and the enthalpy to Joules.

$\displaystyle 25^oC+273=298K$

$\displaystyle -20kJ*\frac{1000J}{1kJ}=-20000J$

Use the given values to solve for the Gibb's free energy.

$\displaystyle \Delta G = (-20,000J) - (298K)(-80\frac{J}{K})$

$\displaystyle \Delta G = 3.84kJ$

Since $\displaystyle \Delta G$ is positive at these conditions, we can conclude that the reaction is nonspontaneous and will not take place.

The sign of the change in Gibb's free energy, $\displaystyle \Delta G$ tells us whether a reaction is spontaneous or not (whether the reaction requires the net input of energy or not). For a spontaneous reaction, the $\displaystyle \Delta G$ value is always less than zero. This is because the free energy of the products is less than that of the reactants. This indicates that there is a net release of energy, as opposed to a net energy consumption. When the products are at a lower energy than the reactants, the reaction can proceed spontaneously, and is known as exothermic.

$\displaystyle \Delta G=G_{\text{products}}-G_{\text{reactants}}$

Gibbs free energy is used to determine if a reaction will occur spontaneously. If the Gibbs free energy is negative for the given process, then the reaction will occur spontaneously.

In the given scenario, ice will always melt spontaneously when the temperature is above 0^oC. As a result, we can state that the Gibbs free energy will always be negative for this process.

$\displaystyle \Delta G = \Delta H - T \Delta S$.

If a process is endothermic, $\displaystyle \Delta H$ is positive.

If a process involves a decrease in entropy, $\displaystyle \Delta S$ is negative.

Under these conditions, the only way $\displaystyle \Delta G$ can be negative is if $\displaystyle T$ is below absolute zero, which is impossible. Catalysts do not effect $\displaystyle \Delta G$, rather they only affect reaction kinetics, not thermodynamics.

There are a few things that can speed up the rate of a reaction. You can increase the temperature or reactant concentrations. You can also add a catalyst to the reaction. Increasing temperature will ensure that reactant molecules move more quickly in solution, causing more collisions and faster reactions. Essentially, increasing temperature will increase the energy available to fuel the reaction. Adding a catalyst decreases the activation energy of the reaction, meaning that the reaction requires less work. Increasing reactant concentrations will shift equilibrium toward the products via Le Chatelier's principle, causing the reaction to speed up in that direction.

Slowly pouring the reactant into the solution will not increase the reaction rate, and can potentially slow it down.

Catalysts and enzymes (biological catalysts) function to increase the rate at which a reaction takes place. They do not affect the amount of products created once at equilibrium. In other words, catalysts will affect kinetics, but not equilibrium. Equilibrium can only be affected by Le Chatelier principles, including temperature change, reactant/product concentration change, or pressure change (for a system involving gases).

Catalysts act by lowering the activation energy of a reaction and stabilizing the transition states. They are not consumed or produced by the reaction, meaning that they are not included in the net reaction equation. Both the forward and reverse reactions will be affected. When the forward reaction is favored by equilibium principles, the catalyst helps make products. When the reverse reaction is favored, the catalyst will still function to help create reactants.

We can compare the rates by setting up the following equations:

$\displaystyle R_1=k[A]^4[B]^3$

$\displaystyle R_2=k[2A]^4[2B]^3$

We can expand the second equation by distributing the exponent:

$\displaystyle R_2=k16[A]^48[B]^3$

$\displaystyle R_2=(16)(8)k[A]^4[B]^3$

$\displaystyle R_2=128k[A]^4[B]^3$

We can see that the rate is multiplied by a factor of 128 when the concentration of each reactant is doubled.