The entropy cannot decrease in an isolated system because the energy can only be degraded. Since the system is isolated, no higher-grade energy—or any energy at all—is being introduced into the system. As a result, the entropy cannot decrease. The other answer choices relate to the other laws of thermodynamics. 

Standard states are defined as a specific set of conditions, such as when a gas is at \(\displaystyle 1.0 atm\), \(\displaystyle 1.0 M\) concentration, and \(\displaystyle 25^{\circ} C\).  

Standard enthalpy of formation, the energy required for form 1 mole of a compound from its constituent elements, occurs when elements are in their standard states. 

\(\displaystyle Q=mc\Delta T Q = (211g) (4.18 J\cdot g^{-1}\cdot ^{\circ}C^{-1})(100^{\circ}C - 13^{\circ}C)= 7.76 \cdot10^{4} J\)

\(\displaystyle Q\) is positive because heat flows into the system to raise the temperature of the water.

\(\displaystyle Q=mc\Delta T = (500g)(4.18 Jg^{-1}^{\circ}C^{-1})(99^{\circ}C - 1^{\circ}C)=2.04\cdot10^{5} J\)

There are four main laws of thermodynamics, which describe how temperature, energy, and entropy behave under various circumstances. The zeroth law of thermodynamics helps to define temperature; it states that if two systems are each in thermal equilibrium with a third system, they must be in thermal equilibrium with each other. The first law of thermodynamics negates the possibility of perpetual motion; it states that when energy passes into or out of a system, the system's internal energy changes in accord with the law of conservation of energy. The second law of thermodynamics also negates the possibility of perpetual motion; it states that in a natural thermodynamic process, the sum of the entropies of the interacting systems increases. Lastly, the third law of thermodynamics states that the entropy of a system approaches a constant value as the temperature nears absolute zero.

Use the equation that relates heat, mass, specific heat, and change in temperature:

\(\displaystyle q=mC\Delta T\)

\(\displaystyle \Delta T=T_{f}-T_{i}=43C-(-12C)=55C\)

\(\displaystyle q=2.5g*.385\frac{J}{g*C}*55C=52.94J\)

A reaction is spontaneous if and only if \(\displaystyle \Delta G^{\circ} < 0\). The sign of \(\displaystyle \Delta H\) only tells us if a reaction is exothermic or endothermic. The rate constant only tells us the rate at which a given chemical reaction proceeds based on the reactants and the products. The first law of thermodynamics only tells us that energy is conserved, and can neither be created nor destroyed. The second law of thermodynamics states that the entropy of the universe is spontaneously increasing, however, under the right conditions, these reactions may proceed.

In this question, we're told of a reaction that is run at a given temperature and with a certain equilibrium constant. We're asked to evaluate the standard change in free energy for this reaction.

First, right off the bat, we can see that the equilibrium constant for the reaction is greater than one. As a result, we know that under standard conditions, equilibrium of this reaction will favor products. Hence, we know right away that \(\displaystyle \Delta G^{o}\) will be negative, which allows us to rule out any answer choices that are positive.

To actually calculate the value of \(\displaystyle \Delta G^{o}\), we'll need to use an equation that relates this quantity with temperature and the equilibrium constant. We can do so by using the following expression.

\(\displaystyle \Delta G^{o}=-RTln(K_{eq})\)

If we plug in the values that we know, we can solve for the correct answer.

\(\displaystyle \Delta G^{o}=-(8.314\: \frac{J}{mol\cdot K})(45+273.15\: K)ln(1.4)\)

\(\displaystyle \Delta G^{o}=-890\: \frac{J}{mol}\)

Recall how to find the change in free energy for a reaction from using free energies of formation:

\(\displaystyle \Delta G^{\degree}_{rxn}=\Sigma n\Delta G^{\degree}_f \text{ of products}-\Sigma n\Delta G^{\degree}_f \text{ of reactants}\) 

Where \(\displaystyle n\) is the number of moles of the molecule.

Plug in the given values to find the change in free energy for the reaction.

\(\displaystyle \Delta G^{\degree}_{rxn}=4(-228.6)-2(159.4)-2(51.3)=-1335.8kJ\)

For this question, we're asked to determine the temperature needed for a reaction to be in a state of equilibrium. We are also provided with the entropy and enthalpy changes for this reaction.

Remember that for a reaction to be in a state of equilibrium, the value for its change in free energy must be equal to zero. Furthermore, we can state the relationship between the change in free energy with the change in entropy, the change in enthalpy, and the temperature as follows.

\(\displaystyle \Delta G=\Delta H-T\Delta S\)

Because the \(\displaystyle \Delta G\) term in the above expression must be equal to zero, we can plug this value in and then isolate the term for temperature.

\(\displaystyle 0=\Delta H-T\Delta S\)

\(\displaystyle T\Delta S=\Delta H\)

\(\displaystyle T=\frac{\Delta H}{\Delta S}\)

Now that we have this expression, we just need to plug in the two values given to us in the question stem. Remember to make the units match though!

\(\displaystyle T=\frac{175\:\frac{kJ}{mol}}{0.5\:\frac{kJ}{mol\cdot K}}=350\:K\)