Download Practice Test 2 Solutions Oct 2010 - University of KwaZulu

Time: 45 Minutes UNIVERSITY OF KWAZULU‐NATAL SCHOOL OF CHEMISTRY HOWARD COLLEGE CAMPUS PHYSICAL CHEMISTRY FOR CHEMICAL ENGINEERS (CHEM 251) PRACTICE TEST 2 ANSWER ALL QUESTIONS Marks: [50] The heat capacity of chloroform (trichloromethane, CHCl3) in the range 240 K to 330 K is given by Cp,m/(J K−1 mol−1) = 91.47 + 7.5 × 10−2 (T/K). In a particular experiment, 5.23 mol CHCl3 is heated from 315.21K to 512.13 K. Calculate the change in molar entropy of the sample. 1.
[10] Answer: ΔS ≈ nC P , m ln
Tf
Ti
Cp,m/(J K−1 mol−1) = 91.47 + 7.5 × 10−2 (330). = 116.22 ∆
5.23
116.22
ln
330
315.21
27.87120363
Data needed (Since this is a take home practice test, additional data can be sourced via the internet, or your book; in a real test the data would be provided in the data section): •
•
•
334.35 K boiling point Cp,m/(J K−1 mol−1) for the gas = 65.33 Since the substance boils at 334.45 K, from 330 – 334.45 K, we will assume over this temperature range that the Cp,m/(J K−1 mol−1) expression given still applies (of course you could use the value for Cp,m of CHCl3 liquid, from a suitable source) From 330 – 334.35 K: ∆
5.23
91.47
7.5
10
2 334.45
ln
334.35
330
7.982308584 Note: Since we assumed that the Cp,m/(J K−1 mol−1) expressions holds to the boiling point, you can calculate the ΔS change from 315.21 K to the boiling point. The difference in ΔS (±0.0782) using both methods is within the error limits of the calculations Next step is the ΔS change from 334.45 K to 512.13 K: ∆
5.23
65.33
ln
512.13
334.35
145.6872384 Change in ΔS = 27.87120363 + 7.982308584 + 145.6872384 = 181.5407507 Answer = 181.54 J K−1 mol−1 2.
Consider a perfect gas contained in a cylinder and separated by a frictionless adiabatic piston into two sections A and B. All changes in B is isothermal; that is, a thermostat surrounds B to keep its temperature constant. There is 3.01 mol of the gas in each section. Initially, TA = TB = 422 K,VA = VB = 12.11 dm3. Energy is supplied as heat to Section A and the piston moves to the right reversibly until the final volume of Section B is 4.05 dm3. Calculate (a) ∆SA and ∆SB,(b) ∆AA and ∆AB,(c) ∆GA and ∆GB,(d) ∆S of the total system and its surroundings. If numerical values cannot be obtained, indicate whether the values should be positive, negative, or zero or are indeterminate from the information given. (Assume CV,m = 20 J K−1 mol−1.) [15] For the entropy problems, calculate it as separate steps, for A, i.e. look at the change in volume, and
then the change in temperature. For B, since it’s an isothermal compression, can simply just
calculate the one process change in terms of the pressure
ΔS = ∫
V
∂qreversible 1
= qreversible = nR ln f T
T
Vi
ΔS ≈ nCV ,m ln
Tf
ΔS ≈ nCP ,m ln
Tf
T
T
The ‘ΔA’ and ‘ΔG’ cannot be determined for ‘A’ compartment For B compartment, (recall test 1, were for the reversible compression of a gas ΔH = ΔU = 0) ΔG = ΔA = TΔS Answer Table for question 2 TA=TB = 422 K VA=VB = 12.11 dm3 nA=nB = 3.01 mol VB2 = 4.05 dm3 VA2 = 20.17 dm3 R = 8.314472 J K−1 mol−1 dm3 bar K‐1 mol‐1 K K bar Bar Bar R = TA2 = TB2 = PA1 = PB1 = PB2 (PA 2)= 0.0831447
2101.664198
422
8.721062001
8.721062001
26.07705206
ΔSA (Tot) 109.4175343 J K−1 ΔSA (step for increase in Temp) 96.64986415 J K−1 ΔSA (Step for increase in Volume) Cv,m (3/2R) = Cp,m (R+Cv,m) = ΔSB (const T) 12.7676702 J K−1 20
28.314472
‐27.41195926 J K−1 ΔAB (U+TΔS) 11567.84681 J mol−1 ΔGB (H+TΔS) 11567.84681 J mol−1 SA + SB = 82.00557509 J K−1 Note: • The B side is isothermal, thus we know that before and after B gets compressed the temperature remains the same • For the B side we know moles (n), new volume (VB2), temperature (TB2), and so we can find pressure (PB2) after the compression using PV = nRT • After the compression, the pressure in B must equal the pressure in A (otherwise the frictionless piston would push back), so now we have pressure in A (PA2), we know the moles (n) in A, we can find the volume in A (initial volume of A 12.11 + the volume displaced in B (12.11 – 4.05)), so solve for T using PV = nRT • ΔSA (Tot) is found by finding the entropy change when the gas is heated from the initial temperature to the final temperature; and when the gas expands from its initial volume to the final volume (see example in the notes/slides about Argon & other examples in TUT problems about calculating S step by step for multi‐event processes). • ΔSB (Tot) is found only from the compression step (nRln(Vf/Vi)) since no change in temperature −1
−1
Total ΔS for the system = ΔSA + ΔSB = ‐80.4 J K mol −1
−1
Recall ΔSsystem = ‐ ΔSsurroundings; so ΔSsurroundings = + 80.4 J K mol 3.
Calculate ∆rG (375 K) for the reaction 2 CO(g) + O2(g) → 2 CO2(g) from the value of ∆rG (298 K),∆rH
(298 K), and the Gibbs–Helmholtz equation. [5] ∆
2
394.36
2
137.17
1
0
514.38 ∆
2
393.51
2
110.53
1
0
565.96 H
⎛ ∂ G⎞
⎟ =− 2
⎜
T
⎝ ∂T T ⎠ P
Or in a more practical form:
∆
375
514.38
298
565.96
1
375
1
298
1.336139418 Answer = ‐501.05 kJ mol‐1 4.
The dissociation vapour pressure of NH4Cl at 427°C is 608 kPa but at 459°C it has risen to 1115 kPa. Calculate (a) the equilibrium constant, (b) the standard reaction Gibbs energy, (c) the standard enthalpy, (d) the standard entropy of dissociation, all at 427°C. Assume that the vapour behaves as a perfect gas and that ∆H and ∆S are independent of temperature in the range given. [10] First step, need the balanced equation for the dissociation process: The “practical” equilibrium expression Kp: Since in practice, one usually measures the concentrations, pressures, volumes, etc directly, thus when looking at the ratios of products over reactants we are usually getting some kind of expression that is related to the units we have used to measure our process. However, the equilibrium constant (dimensionless) is given by the activities of each species: The activity for any pure solid is 1, thus K simplifies to: Activity of any gas is given by (partial pressure of the species (j) divided by the standard pressure (Po = 1bar) Thus when we substitute into the K expression we get: Note: Total pressure: From the balanced equation 1 mol of each product is produced, thus an assumption that is valid in this example is that the partial pressures of each component are equal, thus ~
2
Looking at the K expression once more: 1
4
Thus at 427 °C: 1
4
608
100
9.24160 1115
100
31.08063 At 459 °C 1
4
To find ΔG (at 427 °C) simply use: ∆
ln Thus ΔG = ‐12.9450848 KJ mol‐1 The problem states ΔH doesn’t vary with temperature over the range studies thus to find ΔH (at 427 °C) use ln K 2 − ln K1 = −
Δr H o
R
⎛1 1⎞
⎜⎜ − ⎟⎟ ⎝ T2 T1 ⎠
Thus ΔH = + 161.5436788 KJ mol‐1 The entropy can be found by re – arranging: ∆
∆
∆ Thus ΔS = 249.2162589 J K‐1 mol‐1 (a)
(b)
(c)
(d)
Answer: 9.241; 31.08 Answer: ΔG = ‐12.94 KJ mol‐1 Answer: ΔH = + 161.5 KJ mol‐1 Answer: ΔS = 249.2 J K‐1 mol‐1 5.
Consider the cell Pt|H2(g,p )|HCl(aq)|AgCl(s)|Ag, for which the cell reaction is 2 AgCl(s) + H2(g) → 2 Ag(s) + 2 HCl(aq). At 25°C and a molality of HCl of 0.010 mol kg−1, E = +0.4658 V. (a) Write the Nernst equation for the cell reaction. (b) Calculate ∆rG for the cell reaction. (c) calculate E (AgCl, Ag) [10] Start with the general expression for the Nernst equation (use Q, cause not at equilibrium): ln From the balanced equation given Q: Since activities for solids are 1, & for H2 gas, and in terms of molality (a = γm). ln
=
ln
(b) To find ∆rG use: Δ r G = −vFE ‐1 Thus ∆rG = 2 * 96485 * 0.4658 = 89885.426 J mol‐1 = 89.885 kJ mol
(c) Use the expression from (a) ln
Note (γ = 0.06) would be provided in the data section of an actual test Eo = 0.275 V