Gen CHM II CH 16 Thermodynamics

Of the following, which is not true of a system at equilibrium?

K=eRTΔG∘ For a system at equilibrium, Q=K and ΔG=0. Based on the equation ΔG=ΔG∘+RTlnQ, this implies that 0=ΔG∘+RTlnK. This equation is equivalent to ΔG∘=−RTlnK.

When heat transfer occurs, the heat transferred to/from the system is always:

equal to the heat transferred to/from the surroundings in magnitude, and opposite in sign

The change in Gibbs free energy may be equated to:

-TΔSuniv The change in Gibbs free energy for a process is equal to negative value of the temperature times the corresponding change in entropy for the universe.

Consider the process of ice melting: H2O(s)⟶H2O(l) The entropy change is 22.1 J/K and requires that the surroundings transfer 6.00 kJ of heat to the system. Is the process spontaneous at −10.00∘C? Is it spontaneous at +10.00∘C?

At −10.00∘C(263.15 K), the following is true: ΔSuniv=ΔSsys+ΔSsurr=ΔSsys+qsurrT=22.1JK+−6.00×103 J263.15 K=−0.7 J/K Suniv<0, so melting is nonspontaneous at −10.0∘C. At 10.00∘C(283.15 K) the following is true: ΔSuniv=ΔSsys+ΔSsurr=ΔSsys+qsurrT=22.1JK+−6.00×103 J283.15 K=+0.9 J/K Suniv>0, so melting is spontaneous at 10.0∘C.

Which of the following four equilibrium constant values indicates the greatest (largest positive) free energy change at equilibrium?

Keq=0.15 The free energy change at equilibrium is positive for values of K less than 1, and the equation ΔG=−RTlnK indicates that the smaller the absolute value of K for a reaction, the greater (more positive) the value of ΔG for that reaction. So, the reaction with the largest numerical free energy change is the one in which K=0.15.

The term standard free energy of formation best fits which of the following descriptions?

Standard free energy of formation is the change in free energy accompanying the formation of one mole of a substance from its constituent elements in their standard states.

The temperature difference between the objects is infinitesimally small, Tsys≈Tsurr, and so the heat flow is

Thermodynamically reversible. In this case, the system and surroundings experience entropy changes that are equal in magnitude and therefore sum to yield a value of zero for ΔSuniv. This process involves no change in the entropy of the universe.

The objects are at different temperatures, and heat flows from the hotter to the cooler object.

This is always observed to occur spontaneously. This process involves an increase in the entropy of the universe. ΔSsys=−qrevTsys and ΔSsurr=qrevTsurr

The objects are at different temperatures, and heat flows from the cooler to the hotter object.

This is never observed to occur spontaneously. This process involves a decrease in the entropy of the universe. ΔSsys=qrevTsys and ΔSsurr=−qrevTsurr

For a spontaneous process, which of the following MUST be true? (select all that apply)

TΔS>ΔH ΔSuniv>0 A spontaneous process is associated with a negative value of ΔG (ΔG<0), which by the equation ΔG=ΔH−TΔS must mean that TΔS>ΔH. Recall that spontaneous processes are also ones in which the entropy of the universe increases: ΔSuniv>0.

When calculating the change in free energy using standard free energies of formation, one will need:

When using this method to calculate the change in free energy, we can obtain ΔG simply from the standard free energies of the products and reactants, as long as we know their coefficients in the reaction.

The entropy of any substance at any temperature above absolute zero is called the:

absolute entropy

When a pure, perfect crystalline substance initially at 0 K begins to change its shape or state, its entropy:

increases The entropy of a crystalline solid at absolute zero is zero. It has no kinetic energy, so a change in state or shape can only involve an increase in kinetic energy as its particles begin to move and take on more microstates. Therefore, its entropy can only increase.

In most situations where heat flows between a system and its larger surroundings, the change in entropy for the system is:

less than the change in entropy for the surroundings

Any process with a positive change in enthalpy and a negative change in entropy will be:

nonspontaneous

The third law of thermodynamics describes the entropy of a

perfect crystalline solid.

The absolute entropy for any substance will always be:

positive If the entropy of a crystalline solid at absolute zero is zero, then its entropy can only increase with increasing heat energy from that point as the particles begin to move and take on more microstates. Thus the absolute entropy of a substance must always be a positive value, regardless of the temperature or the identity of the substance.

A process will definitely be spontaneous if:

the entropy of the universe increases in the process When the entropy of a system increases, this may or may not mean that a particular process is spontaneous. But if the entropy of the universe increases due to a particular process, that process will certainly be spontaneous.

If K=1

ΔG = 0: reactants and products are equally abundant

If K< 1, __________.

ΔG is positive: reactants are more abundant

If K>1, _______.

ΔG∘ is negative: products are more abundant

Water melting at a certain temperature is a spontaneous process if:

ΔSuniv>0 The second law of thermodynamics states that a process is spontaneous if, for that process, ΔSuniv>0.

Calculate the standard free energy change for the reaction below. P4(s)+6Cl2(g)→4PCl3(g) ΔG∘298,P4(s)ΔG∘298,PCl3(g)=24.44 kJ/mol=−267.8 kJ/mol

−1095.6 kJ/mol Note that the free energy of formation of Cl2(g) at 298 K is 0 kJ/mol, as this is a pure element in its standard state. Bearing this in mind, and using the free energies of formation provided, we obtain the following: ΔG∘298=∑νΔG∘f,products−∑νΔG∘f,reactants=4ΔG∘298,PCl3(g)−[ΔG∘298,P4(s)+6ΔG∘298,Cl2(g)]=4(−267.8 kJ/mol)−[(24.44 kJ/mol)+6(0 kJ/mol)]=−1095.6 kJ/mol