Homework Problem Set 8 Solutions

Transcription

1 Chemistry 360 Dr. Jean M. Standard Homework roblem Set 8 Solutions. Starting from G = H S, derive the fundamental equation for G. o begin, we take the differential of G, dg = dh d( S) = dh ds Sd. Next, we can substitute the fundamental equation for H, dh = ds + d, dg = dh ds Sd = ds + d ds Sd dg = d Sd. hus, we see that the natural variables of G are and.. Calculate the entropy change when neon at 5 C and atm is allowed to expand from L to.50 L. he system is simultaneously heated to 85 C. Assume ideal behavior. [Hint: use one of the fundamental equations to obtain an expression for the differential ds and integrate.] We can use the fundamental equation for U to derive an expression to use in this case for calculating the entropy change. he fundamental equation is du = ds d. Rearranging to isolate the term involving entropy, we have Dividing both sides by gives, ds = du + d. ds = du + d. Next, for an ideal gas, we know that du = C v d. Substituting, we have ds = C v d + d. Finally, for an ideal gas, we can use the relation that = nr. Substituting, ds = C v d + nr d.

2 . continued Integrating, ΔS = C v d + nr d = C v ln + nrln. Neon is a monatomic ideal gas and so its constant volume heat capacity is C v = 3 nr. he expression for the entropy change therefore becomes ΔS = C v ln + nrln = 3 nrln + nrln. he number of moles of neon must be determined in order to complete the calculation. Assuming ideal gas behavior, the number of moles may be calculated from the initial conditions, Substituting, ΔS = 3 nrln + nrln n = R ( atm) L = Latm/molK ( ) ( )( 98.5K) n = mol. = 3 ( mol ) 8.34Jmol K ΔS = 0.0 J/K. ( )ln 358.5K + ( mol) ( 8.34Jmol K )ln.50l 98.5K L

3 3. One mole of H gas undergoes the following transformation: 3 H (g) ( 00 K, bar) H (g) ( 500 K, 0 bar). he molar constant pressure heat capacity of H is 0.0 J/mol-K. Determine ΔS for this process, assuming H behaves ideally. his time, we can use the fundamental equation for H to derive an expression to use for calculating the entropy change. he fundamental equation is dh = ds + d. Rearranging to isolate the term involving entropy, we have Dividing both sides by gives, ds = dh d. ds = dh d. Next, for an ideal gas, we know that dh = C p d. Substituting, we have ds = C p d d. Finally, for an ideal gas, we can use the relation that = nr. Substituting, Integrating, ΔS = ds = C p d nr d. C p d nr d = C p ln nrln. he molar constant pressure heat capacity of H is given as is C p,m = 0.0 J mol - K -. We can rewrite the result above in terms of C p,m as C p = nc p,m. he expression for the entropy change therefore becomes ΔS = nc p ln nrln.

4 3. continued 4 Substituting, ΔS = nc p ln nrln = ( mol) 0.0 J mol K ΔS = 3.0 J/K. ( )ln 500K 00K ( )ln 0bar bar ( mol) 8.34Jmol K

5 4. Determine ΔU and ΔA for a process in which 0.60 mol of an ideal gas expands reversibly and isothermally from 0.0 L to 5.0 L at 37 C. 5 For the determination of ΔU, we note that for an ideal gas, we have the relation du = C v d. Since the temperature is constant, d = 0, so the equation becomes du = 0, or ΔU = 0. For the determination of ΔA, a good place to start is the fundamental equation for A, da = S d d. Since the temperature is constant, d = 0, so the equation becomes Integrating yields da = d. da = d. Since the system is an ideal gas, we can substitute the ideal gas relation for, nr ΔA = d. he process is isothermal, so the factor nr can be pulled out and the integration performed, Substituting, ΔA = nr ln nr ΔA = d = nr d % ΔA = nr ln ( *. & ) = ( 0.60mol) ( 8.34Jmol K ) 30.5K ΔA = 378J. ( )ln 5.0L 0.0L

6 5. A sample of mol of argon gas expands from.0 L to 5.0 L at a constant temperature of 500 K. Determine the quantities ΔH, ΔS, and ΔG for this process. 6 For the determination of ΔH, we note that for an ideal gas, we have the relation dh = C p d. Since the temperature is constant, d = 0, so the equation becomes dh = 0, or ΔH = 0. For the determination of ΔG, a good place to start is the fundamental equation for G, dg = S d + d. Since the temperature is constant, d = 0, and the equation becomes dg = d. Integrating yields dg = d. Since the system is an ideal gas, we can substitute the ideal gas relation for, ΔG = nr d. he process is isothermal, so the factor nr can be pulled out and the integration performed, nr ΔG = d = nr d $ ΔG = nr ln & ). % ( Note that we do not know the change in pressure, but we do know the change in volume. At constant temperature, =. herefore, =, and the Gibbs free energy change becomes ΔG = nr ln &. $

7 5. continued 7 Substituting, ΔG = nr ln = ( 0.988mol) ( 8.34Jmol K ) 500 K ΔG = 3760 J. ( )ln.0l 5.0L Finally, to determine the entropy change, we can use the defition of the Gibss Free Energy, G = H S. Converting this to the change in G, ΔG = ΔH Δ( S) = ΔH ΔS SΔ. Since the temperature is constant, Δ = 0, and the last term drops out, yielding Solving for the entropy change ΔS, we have ΔG = ΔH ΔS. ΔS = ΔH ΔG. Substituting, ΔS = ΔH ΔG = K 3760 J 500 K ΔS = 7.5 J/K.

8 6. hree moles of toluene are vaporized at the boiling point of C. he heat of vaporization is 36.9 J/g. Calculate ΔH, ΔS, and ΔG for this phase transition. 8 he molar enthalpy of vaporization can be determined using the molecular weight of toluene, which is 9.4 g/mol, ΔH m = 36.9 J/g % $ ( ) 9.4 g mol ΔH m = J/mol (or 33.35kJ/mol). o determine the enthalpy change for three moles of toluene, we use the relation Substituting, we have ΔH = nδh m. & ( ΔH = nδh m = ( 3 mol) ( 33.35kJ/mol) ΔH = 00. kj. For a phase transition, ΔS = ΔH. herefore, the entropy of vaporization is ΔS vap = = ΔH vap J 384.5K ΔS vap = 60.4JK. he Gibbs free energy at constant temperature is ΔG = ΔH ΔS. Substituting the expression for the entropy yields ΔG vap = ΔH vap ΔS vap = ΔH vap ΔH vap = ΔH vap ΔH vap ΔG vap = 0.

9 7. Liquid water is compressed isothermally from bar to 000 bar. Calculate the change in the molar Gibbs free energy. Assume that the molar volume is independent of pressure. 9 he best place to start is from the fundamental equation for the Gibbs free energy, dg = S d + d. At constant temperature, d = 0, so dg = d. In terms of molar quantities, dg m = m d. Integrating, ΔG m = m d. If the molar volume is constant (independent of pressure), then it can be pulled out of the integral to yield ΔG m = m d = m d ΔG m = m ( ). In order to evaluate the change in molar Gibbs free energy, we need the molar volume of liquid water. his can be calculated from the inverse of the density ( g/ml), Substituting, " m = ml %" $ 8.0 g %" L % $ $.0 g & mol & 000 ml & m = L/mol. ( ) ( )( 000 bar bar) ΔG m = m = L/mol = 8.0 L bar mol $ 00 J & ) % L bar ( ΔG m = 800 J/mol.

10 8. Using information from the appendix of the text, calculate ΔG R at 5 C for the reaction 0 CO( g) + O ( g) CO ( g). Is the reaction spontaneous? he standard molar Gibbs free energies of formation at 5 C as obtained from the appendix are: Compound ΔG f (kj/mol) CO (g) 37. O (g) 0 CO (g) he standard molar Gibbs free energy of reaction is given by ΔG R ΔG R = ΔG f ( CO ) ΔG f CO ( ) ΔG f ( O ) ( ) ( 37. kj/mol) 0 kj/mol ( ) = kj/mol = 57. kj/mol. he reaction is spontaneous since the standard molar Gibbs free energy is negative. 9. Determine ΔG R of the text, at 5 C for the following hydrogenation reaction using information from the appendix C 6 H 6 ( l) + 3H ( g) C 6 H ( l). Is the reaction spontaneous? he standard molar Gibbs free energies of formation at 5 C as obtained from the appendix are: Compound ΔG f (kj/mol) C 6 H 6 (l) 4.3 H (g) 0 C 6 H (l) 6.8 he standard molar Gibbs free energy of reaction is given by ΔG R ΔG R = ΔG f ( C 6 H ) ΔG f C 6 H 6 ( ) 3ΔG f ( H ) ( ) ( 4.3 kj/mol) 3( 0 kj/mol) = 6.8kJ/mol = 97.5 kj/mol. he reaction is spontaneous since the standard molar Gibbs free energy is negative.

11 0. Calculate the standard molar Gibbs free energy of reaction at 500 K for the reaction N ( g) + 3H ( g) NH 3 ( g). Use as a starting point values of the standard molar enthalpies and Gibbs free energies of formation at 5ºC, which are available in the appendix of your textbook. he standard molar Gibbs free energy at another temperature may be estimated using the Gibbs-Helmholtz equation, ( ) / $ ΔG / & ) ) ( = ΔH. If we approximate the partial derivative by finite changes, the Gibbs-Helmholtz equation becomes ( ) Δ ( / ) Δ ΔG / ΔH, $ or Δ ΔG & % ( ) ΔH Δ $ & % ). ( he changes are defined as the differences in the higher temperature (500 K in this case) and 98 K, Δ & = $ f i and Δ ΔG & = ΔG ΔG. $ =f =i Inserting these relations into the approximate Gibbs-Helmholtz equation, Δ ΔG & % $ ( ΔH Δ & % $ ( ΔG ΔG ΔH =f % =i $ f & (. i Substituting the temperatures, ΔG ΔG % ΔH =f =i & f ( * i ) ΔG K ΔG K ΔH % K ( *. & 98 K )

12 0. Continued Here, it has been assumed as indicated in the problem that the enthalpy of reaction is constant and the value at 98 K (available in tables) has been used. Isolating the unknown Gibbs free energy change at 500 K, we have ΔG K ΔG K + ΔH % K ( *, & 98 K ) + or ΔG 500 ( 500 K) ΔG K + ΔH % K (. - * 0., & 98 K )/ In order to complete the determination of the Gibbs free energy change at 500K, the values of the Gibbs free energy and the enthalpy of reaction at 98 K are required. hese can be obtained from the tables in the appendix. he standard molar enthalpies and Gibbs free energies of formation at 5 C of the reactants and products are: Compound ΔH f (kj/mol) ΔG f (kj/mol) N (g) 0 0 H (g) 0 0 NH 3 (g) he standard molar enthalpy of reaction at 5 C is ΔH R ΔH R = ΔH f ( NH 3 ) ΔH f ( N ) 3ΔH f ( H ) ( ) ( 0 kj/mol) 3( 0 kj/mol) = 46.kJ/mol = 9. kj/mol. he standard molar Gibbs free energy of reaction at 5 C is ΔG R ΔG R = ΔG f ( NH 3 ) ΔG f ( N ) 3ΔG f ( H ) ( ) ( 0 kj/mol) 3( 0 kj/mol) = 6.5kJ/mol = 33.0 kj/mol. Now we can substitute these values into the expression above to get the Gibbs free energy of reaction at 500 K, + ΔG 500 ( 500 K) ΔG K + ΔH % K (. - * 0, & 98 K )/ kj/mol ( 500 K) -, 98 K ΔG kj/mol. % + ( 9. kj/mol) 500 K (. * 0 & 98 K )/ his is a pretty big change from the value at 98 K and indicates a change in spontaneity at higher temperatures (due to the change in sign of ΔG from 98 to 500 K).

13 . Consider a gas that obeys the equation of state given by 3 Z = + B m. (a) Determine an expression for U & $. Since a partial derivative of U is requested, it is helpful to start with the fundamental equation for U, du = ds d. ake the partial derivative with respect to (at constant ) of both sides of the equation, U & $ = S & $. he partial derivative on the right side of the equation cannot be determined from the equation of state. However, one of the Maxwell relations can be used, S & $ = & $. Substituting yields U & $ = & $. Now, the partial derivative on the right side can be calculated from the equation of state. Since the equation of state is Z = + B m, solving for gives m R = + or = R m B m, + BR m. Now, converting to instead of m leads to the expression = nr + n BR. aking the partial derivative, & $ = nr + n BR.

14 (a). Continued 4 Substituting into the expression for the partial derivative of U gives the result, U & $ U & $ = & $ = nr % $ = % $ nr = = 0. + n BR + n BR & ( & ( (b) Determine ΔS for the isothermal compression at 500 K from 00 L to 0 L of two moles of the gas. he constant B for the gas equals 0. L/mol. he definition of the entropy is ds = dq rev. From the first law, du = dq + dw, or for a reversible process, du = dq rev d. Solving for dq rev yields dq rev = du + d. he exact differential for U is given by du = C v d + U d. We showed that U & = 0 for this gas in part (a). Substituting, the expression for dq rev then becomes $ dq rev = C v d + d. Finally, the entropy is given by ds = C v d + d. For an isothermal process, d = 0, and so the entropy becomes ds = d.

15 (b). Continued 5 his expression can be integrated if / for this gas can be evaluated. From the equation of state, or m R Z = + = + B m, B m. Solving for /, we get or = R + m = nr Substituting into the expression for entropy leads to BR m, + n BR. ds = " $ nr + n BR % d. & Breaking this up into two terms and integrating gives ΔS = nr d + n BR d = nrln + n BR + = ( mol) 8.34 J mol K ( )ln 0L 00L + ( mol) ( 0.Lmol )( 8.34 J mol K ) = 38.9 J/K J/K ΔS = 37.9 J/K. 0 L + 00 L Note that the entropy change is negative because the process involves a compression.

16 . Starting from the fundamental equation for the enthalpy H, dh = ds + d, show that 6 H & $ = ( α ), where α is the thermal expansion coefficient, α = $ & ) % ( Starting from the fundamental equation for H, we need to take the partial derivative with respect to (at constant ) of both sides of the equation,. H & $ = S & $ +. he partial derivative on the right side of the equation can be determined from one of the Maxwell relations, S & $ = & $. Substituting yields H & $ = & $ +. he partial derivative & $ is related to the thermal expansion coefficient, & $ = α. Substituting, H & $ H & $ = & $ = α + = ( α ). +