Chem Quiz 10
Why does water spilled on the floor evaporate even though ΔG° for the evaporation process is positive at room temperature?
Even though ∆G°vap is positive at room temperature, liquid water is in equilibrium with water vapor at a pressure of 0.0313 atm. The vapor pressure of water is just not as high as 1 atm, the standard conditions of a gas. This means that some water will evaporate.
Why do exothermic processes tend to be spontaneous at low temperatures? Why does their tendency toward spontaneity decrease with increasing temperature?
Exothermic processes (-deltaH(sys)) tend to be spontaneous at low temperatures because they increase the entropy of the surroundings (+deltaS(surr)). Because deltaS(surr)= -deltaH(sys)/T , as temperature increases, a given negative deltaH produces a smaller positive deltaS(surr); therefore, exothermicity becomes less of a determining factor for spontaneity as temperature increases.
Why is the standard entropy of a substance in the gas state greater than its standard entropy in the liquid state?
Gases have much greater standard entropy because they have more energetically equivalent ways to arrange their components, which in turn results in greater energy dispersal at 25 °C.
What is the significance of the change in Gibbs free energy (ΔG) for a reaction?
The change in Gibbs free energy for a process is proportional to the negative of deltaS(univ). Because deltaS(univ) is a criterion for spontaneity, deltaG is also a criterion for spontaneity (though opposite in sign).
Why is free energy "free"?
The change in free energy of a chemical reaction represents the maximum amount of energy available, or free, to do work (if ∆G°rxn is negative).
How do you calculate the change in free energy for a reaction under nonstandard conditions?
The free energy of reaction under nonstandard conditions (∆Grxn) can be calculated from ∆G°rxn using the relationship ∆Grxn = ∆G°rxn + RT lnQ, where Q is the reaction quotient (defined in Section 14.7), T is the temperature in K, and R is the gas constant in the appropriate units (8.314 J/mol*K).
How does the standard entropy of a substance depend on its molar mass? On its molecular complexity?
The larger the molar mass, the greater the entropy at 25 °C. For a given state of matter, entropy generally increases with increasing molecular complexity.
How does the value of ΔG° for a reaction relate to the equilibrium constant for the reaction? What does a negative ΔG° for a reaction imply about K for the reaction? A positive ΔG°?
The relationship between ∆G°rxn and K is ∆G°rxn = -RT lnK. When ∆G°rxn is positive, the log of K is negative and K < 1. Under standard conditions (when Q = 1), the reaction is spontaneous in the reverse direction. When ∆G°rxn is zero, the log of K is zero and K = 1. The reaction happens to be at equilibrium under standard conditions. When ∆G°rxn is negative, the log of K is positive and K > 1. Under standard conditions (when Q = 1), the reaction is spontaneous in the forward direction.
Explain the difference between ΔG° and ΔG.
The standard free energy change for a reaction (∆G°rxn) applies only to standard conditions. For a gas, standard conditions are those in which the pure gas is present at a partial pressure of 1 atm. For nonstandard conditions, we need to calculate ∆Grxn (not ∆G°rxn) to predict spontaneity.
State the third law of thermodynamics and explain its significance.
The third law of thermodynamics states that the entropy of a perfect crystal at absolute zero (0 K) is zero. For enthalpy, we defined a standard state so that we could define a "zero" for the scale. This is not necessary for entropy because there is an absolute zero.
What are three different methods to calculate ΔG° for a reaction? Which method would you choose to calculate ΔG° for a reaction at a temperature other than 25 °C?
The three ways of calculating the ∆Grxn are: WRITE The method to calculate the free energy of a reaction at temperatures other than at 25 °C is the first method. The second method is only applicable at 25 °C. The third method is only applicable at the temperature of the individual reactions, generally 25 °C.
How can you calculate the standard entropy change for a reaction from tables of standard entropies?
To calculate ,subtract the standard entropies of the reactants multiplied by their stoichiometric co- efficients, from the standard entropies of the products multiplied by their stoichiometric coefficients, or WRITE where n(p) represents the stoichiometric coefficients of the products, n(r) represents the stoichiometric coefficients of the reactants, and S° represents the standard entropies. Keep in mind when using this equation that, unlike enthalpies of formation, which are equal to zero for elements in their standard states, standard entropies are always nonzero at 25 °C.
Predict the spontaneity of a reaction (and the temperature dependence of the spontaneity) for each possible combination of signs for ΔH and ΔS (for the system). a) ΔH negative, ΔS positive b) ΔH positive, ΔS negative c) ΔH negative, ΔS negative d) ΔH positive, ΔS positive
a) change in free energy will be negative at all temperatures and the reaction will therefore be spontaneous at all temperatures b) change in free energy will be positive at all temperatures and the reaction will therefore be nonspontaneous at all temperatures. c) change in free energy will depend on temperature. The reaction will be spontaneous at low temperature but nonspontaneous at high temperature. d) change in free energy will also depend on temperature. The reaction will be nonsponta- neous at low temperature but spontaneous at high temperature.
