The change in entropy is calculated by:

$\displaystyle \Delta S_{products}-\Delta S_{reactants}$

When $\displaystyle \Delta S$ cannot be measured, it can be calculated from known entropies of formation.

It is important to first balance the reaction before performing calculations. The coefficients are important in determining the change in entropy of a reaction.

$\displaystyle N_{2(g)}+2O_{2(g)}\rightarrow 2NO_2$

$\displaystyle \Delta S_{rxn}=2(\Delta S_{NO_2})-\Delta S_{N_{2(g)}}-2(\Delta S_{O_{2(g)}})$

$\displaystyle \Delta S_{rxn}=2(240J)-191.5J-2(205J)$

$\displaystyle \Delta S_{rxn}=-121.5\frac{J}{mol}$

The change in entropy is calculated by:

$\displaystyle \Delta S_{products}-\Delta S_{reactants}$

When $\displaystyle \Delta S$ cannot be measured, it can be calculated from known entropies of formation.

It is important to first balance the reaction before performing calculations. The coefficients are important in determining the change in entropy of a reaction.

$\displaystyle 2Mg +O_2\rightarrow 2MgO$

$\displaystyle \Delta S_{rxn}=2(\Delta S_{MgO})-2(\Delta S_{Mg})-\Delta S_O_2{}$

$\displaystyle \Delta S_{rxn}=2(26.9J)-2(32.7J)-205J$

$\displaystyle \Delta S_{rxn}=-216.6\frac{J}{mol}$

The change in entropy is calculated by:

$\displaystyle \Delta S_{products}-\Delta S_{reactants}$

When $\displaystyle \Delta S$ cannot be measured, it can be calculated from known entropies of formation.

It is important to first balance the reaction before performing calculations. The coefficients are important in determining the change in entropy of a reaction.

$\displaystyle 2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O$

$\displaystyle \Delta S_{rxn}=8(\Delta S_{CO_2})+10(\Delta S_{H_2O})-2(\Delta S_{C_2H_{10}})-13(\Delta S_{O_2})$

$\displaystyle \Delta S_{rxn}=8(213.6J)+10(69.9J)-2(310J)-13(205J)$

$\displaystyle \Delta S_{rxn}=-877.2\frac{J}{mol}$

The change in entropy is calculated by:

$\displaystyle \Delta S_{products}-\Delta S_{reactants}$

When $\displaystyle \Delta S$ cannot be measured, it can be calculated from known entropies of formation.

It is important to first balance the reaction before performing calculations. The coefficients are important in determining the change in entropy of a reaction.

$\displaystyle A+B\rightarrow C+D$

$\displaystyle \Delta S_{rxn}=\Delta S_{C}+\Delta S_{D}-\Delta S_{A}-\Delta S_{B}$

$\displaystyle \Delta S_{rxn}=-75J+(-45J)-(-10J)-15J$

$\displaystyle \Delta S_{rxn}=-125\frac{J}{mol}$

A reaction will have a positive change in entropy when the disorder of the system is increasing. This can happen when a liquid transitions to a gas, or more gas molecules are formed in the reaction.

$\displaystyle 2NO_{2}_{(g)} \rightarrow N_{2}O_{4}_{(g)}$

Since two molecules of gaseous nitrogen dioxide can combine to form one molecule of dinitrogen tetroxide, the amount of gaseous molecules is halved. This increases order, which decreases the entropy of the system.

Equilibrium is the basis of entropy, which can essentially be thought of as the distribution of energy in the system. For a reaction at a given temperature, the system's entropy will be at a maximum when it reaches equilbrium because this is the ideal energy state, and the forward and reverse reactions are occurring at a steady rate.

Since we have the entropies of formation, we can find the entropy of the reaction using the following equation:

$\displaystyle \Delta S_{reaction} = \Delta S_{products} - \Delta S_{reactants}$

We will need to use the coefficients from the balanced equation to calculate the entropy.

$\displaystyle N_{2}_{(g)} + 3H_{2}_{(g)} \rightarrow 2NH_{3}_{(g)}$

$\displaystyle \Delta S_{reaction}=2(\Delta S_{NH_3})-\Delta S_{N_2}-3(\Delta S_{H_2})$

$\displaystyle \Delta S_{reaction}=2(192.5\frac{J}{molK})-(191.49\frac{J}{molK})-3(130.59\frac{J}{molK})$

$\displaystyle \Delta S_{reaction}=-197.77\frac{J}{molK}$

Since this is the entropy for the reaction when two moles of ammonia are formed, we must divide this entropy by two in order to solve for the entropy of one mole.

$\displaystyle \frac{197.77\frac{J}{molK}}{2}=-99\frac{J}{molK}$

Approximately 467 K. ∆G=∆H-T∆S and a rxn proceeds spontaneously when ∆G < 0
and is non-spontaneous when ∆G > 0. So if we set ∆G=0 and solve the equation for T, we will see that the crossover from spontaneous to non-spontaneous occurs when T=467K.

If Q is greater than K, the reaction has exceeded the equilibrium state. It will proceed nonspontaneously (since equilibrium has already been reached), and this must mean that the ΔG (gibbs free energy) must be positive, or greater than zero. 

A reaction is spontaneous if the Gibb's Free Energy of the reaction is negative.

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

If $\displaystyle \Delta H$, the enthalpy, and $\displaystyle \Delta S$, the entropy, are both negative, then the reaction will be spontaneous if and only if the magnitude of the enthalpy is greater than the magnitude of the entropy times the temperature.