Answers to Practice Test Questions 8 Effect of Temperature on Equilibrium. (a)-(c) solid liquid solid critical point liquid gas gas triple point xenon neon (d) The normal boiling point of the noble gas on the left is higher so it must be the larger noble gas (xenon). Find normal boiling point by finding the temperature along the liquid/gas curve at P = bar. The gas on the left (xenon) has a normal boiling point of 65K; the gas on the right (neon) has a normal boiling point of 27K. Note that you cannot compare critical points or triple points because you must compare points at either the same pressure (e.g. normal boiling points are always at atm bar) or same temperature to be able to draw any conclusions about behaviour.
2. P solid liquid critical point triple point gas T Assume a normal phase diagram (as shown above). The substance will begins as a gas. As it is compressed, its density increases until it becomes a liquid. Continued compression increases the density further until the substance becomes a solid. This progression is shown by the red arrow on the phase diagram. 3. While this question can be solved from scratch using activities and equilibrium constant expressions, it s easier to just use Raoult s law (which is derived from those activities and equilibrium constant expressions): pp HH2 OO = XX HH2 OO pp HH2 OO Either way, you will need to calculate the mole fraction of water. Step : Calculate the moles of each species in solution nn HH2 OO = 300 gg = 6.6526 = 6.7 HH 8.052 gg 2OO To calculate the moles of each ion in solution, it is easiest to calculate the moles of (NNNN 4 ) 2 SSSS 4 + then use mole ratios to calculate the moles of NNNN 4 and SSSS 2 4. nn (NNNN4 ) 2 SSSS 4 = 50 gg = 0.38 (NNNN 32.40 gg 4) 2 SSSS 4 nn NNNN4 + = 0.38 (NNNN 4 ) 2 SSSS 4 2 NNNN 4 + = 0.76 NNNN (NNNN 4 ) 2 SSSS 4 + 4 nn SSSS4 2 = 0.38 (NNNN 4 ) 2 SSSS 4 SSSS 4 2 = 0.38 SSSS (NNNN 4 ) 2 SSSS 4 2 4 Step 2: Calculate the mole fraction of water in the solution XX HH2 OO = nn HH 2OO nn HH 2OO + nn NNNN4 + + nn SSSS4 2 = 6.7 (6.7 )+(0.76 )+(0.38 ) = 0.936 Step 3: Calculate the vapour pressure above the solution pp HH2 OO = XX HH2 OO pp HH2 OO = (0.936)(7373 PPPP) = 6902 PPPP = 6.90 0 3 PPPP = 6.90 kk Step 4: Check your work Does your answer seem reasonable? Are sig. fig. correct? The dissolved salt lowers the vapour pressure of water above the solution.
4. The boiling point of benzene at 0.025 bar is the temperature at which the equilibrium vapour pressure of benzene is 0.025 bar (since boiling point is the temperature at which vapour pressure = atmospheric pressure and, here, the atmosphere has a pressure of 0.025). You have been given the normal boiling point of benzene (i.e. the temperature at which the equilibrium vapour pressure of benzene is atm. So, that makes this a two temperatures; two equilibrium constants question which requires the use of. KK RR TT TT 2 The key to these types of questions is to be organized! Make sure that K is the equilibrium constant corresponding to T (and that K2 corresponds to T2)! Step : Write a balanced chemical equation for the reaction (phase change) CC 6 HH 6(ll) CC 6 HH 6(gg) Step 2: Work out how to calculate activities for all reactants and products aa CC6 HH 6(ll) = aa CC6 HH 6(gg) = pp CC6HH 6(gg) bbbbbb Step 3: Write equilibrium constant expression for the reaction (phase change) and simplify where possible. KK = aa CC6HH 6(gg) = aa CC 6HH 6(ll) pp CC 6HH 6(gg) bbbbbb () = pp CC6HH 6(gg) bbbbbb Step 4: Calculate equilibrium constants, and match them to the corresponding temperature TT = 80.0 = 353.25 KK KK = pp CC6HH 6(gg) bbbbbb TT 2 =??? = aaaaaa.0325 bbbbbb =.0325 bbbbbb aaaaaa 0.025 bbbbbb normal boiling point pp CC6 HH 6(gg) = aaaaaa boiling point at 0.025 bar pp CC6 HH 6(gg) = 0.025 bbbbbb KK 2 = pp CC6HH 6(gg) = = 0.025 bbbbbb bbbbbb Step 5: Look up (or calculate) standard enthalpy change for the reaction (phase change) rr HH = +30.7 = 3.07 0 4 provided in question Step 6: Crunch the numbers KK RR TT TT 2 llll 0.025 = 3.07 04.0325 353.25 KK TT 2 8.34 462 KK llll 0.025.0325 8.34 462 KK 3.07 0 4 TT 2 = llll 353.25 KK = 353.25 KK TT 2 0.025.0325 8.34 462 3.07 0 4 KK = 0.002 830 9 0.00 00 TT 2 KK KK = 0.003 83 TT 2 KK TT 2 = = 26 KK = 2 0.003 83 KK
Step 7: Check your work Does your answer seem reasonable? Since 0.025 bar is a much lower pressure than atm, it makes sense that the boiling point at that (lower) pressure is much lower than the normal boiling point. Step 8: Answer the question! There was a verbal question attached to this calculation. It s easy to forget about that. Always re-read the question when you think you re done to make sure you answered everything! Yes, benzene will boil at room temperature when the pressure is 0.025 bar. Its boiling point at this pressure is 2. 5. This is another two temperatures; two equilibrium constants question requiring the use of. In this case, we can use thermodynamic data to calculate the KK RR TT TT 2 equilibrium constant at 25 from the standard free energy change for the reaction. Step : Write a balanced chemical equation for the reaction (phase change) TTTTTTTT 4(ll) TTTTTTTT 4(gg) Step 2: Work out how to calculate activities for all reactants and products aa TTTTTTTT4(ll) = aa TTTTTTTT4(gg) = pp TTTTTTTT 4(gg) bbbbbb Step 3: Write equilibrium constant expression for the reaction (phase change) and simplify where possible. KK = aa TTTTTTTT 4(gg) = aa TTTTTTTT4(ll) pp TTTTTTTT4(gg) bbbbbb () = pp TTTTTTTT 4(gg) bbbbbb Step 4: Calculate the standard free energy change for the reaction (phase change) rr GG = ff GG (pppppppppppppppp) ff GG (rrrrrrrrrrrrrrrrrr) rr GG = ff GG TTTTTTTT 4(gg) ff GG TTTTTTTT 4(ll) rr GG = 726.7 737.2 rr GG = 0.5 Step 5: Calculate equilibrium constants, and match them to the corresponding temperature TT = 25 = 298.5 KK standard conditions rr GG = RRRRRRRRKK llllkk = rrgg RRRR llllkk = llllkk = 4.24 KK = ee 4.24 KK =.4 0 2 0.5 000 8.34 462 KK (298.5 KK)
TT 2 =??? normal boiling point pp TTTTTTTT4(gg) = aaaaaa KK 2 = pp TTTTTTTT 4(gg) bbbbbb = aaaaaa.0325 bbbbbb =.0325 bbbbbb aaaaaa Step 6: Calculate the standard enthalpy change for the reaction (phase change) rr HH = ff HH (pppppppppppppppp) ff HH (rrrrrrrrrrrrrrrrrr) rr HH = ff HH TTTTTTTT 4(gg) ff HH TTTTTTTT 4(ll) rr HH = 763.2 804.2 rr HH = +4.0 = 4.0 04 mmmmll Step 7: Crunch the numbers KK RR TT TT 2 llll.0325.4 0 2 = 4.0 04 8.34 462 KK llll.0325.4 0 2 8.34 462 KK 4.0 0 4 = TT 2 298.5 KK llll 298.5 KK TT 2 = 298.5 KK TT 2.0325.4 0 2 8.34 462 KK 4.0 0 4 TT 2 = 0.003 354 0 KK 0.000 862 KK TT 2 = 0.002 492 KK TT 2 = 0.002492 KK = 40.2 KK = 28. Step 8: Check your work Does your answer seem reasonable? Since TiCl4 is a liquid at room temperature, its normal boiling point should be higher than room temperature. 6. This is another two temperatures; two equilibrium constants question requiring the use of. Again, keeping organized is key to success. KK RR TT TT 2 Because the equilibrium constant was provided directly (rather than indirectly in the form of pressure data), it is not necessary to calculate activities for this question. Step : Write a balanced chemical equation for the reaction This was provided in the question. FFFFFF (ss) + CCCC (gg) FFFF (ss) + CCCC 2(gg) Step 2: Match equilibrium constants to the corresponding temperature TT = 998 KK TT 2 = 298 KK KK = 0.259 KK 2 =???
Step 3: Calculate the standard enthalpy change for the reaction rr HH = ff HH (pppppppppppppppp) ff HH (rrrrrrrrrrrrrrrrrr) rr HH = ff HH FFFF (ss) + ff HH CCCC 2(gg) ff HH FFFFFF (ss) + ff HH CCCC (gg) rr HH = 0 + 393.5 272.0 rr HH =.0 =.0 04 Step 4: Crunch the numbers KK RR TT TT 2 =.0 04 0.259 0.259 = 3. 8.34 462 KK 998 KK 298 KK + 0.5 KK 2 0.259 = ee3. KK 2 0.259 = 23 KK 2 = (0.259)(23) KK 2 = 5.8 Step 5: Check your work Does your answer seem reasonable? This reaction is exothermic ( rr HH < 0), so it is expected to have a larger equilibrium constant at the lower temperature. 7. (a) Equilibrium constants can be calculated from the standard free energy change for a reaction. Step : Write a balanced chemical equation for the reaction This was provided in the question. 3 HH 2(gg) + NN 2(gg) 2 NNNN 3(gg) Step 2: Convert temperature to Kelvin (if necessary) TT = 25 = 298.5 KK Step 3: Calculate the standard free energy change for the reaction rr GG = ff GG (pppppppppppppppp) ff GG (rrrrrrrrrrrrrrrrrr) rr GG = 2 ff GG NNNN 3(gg) 3 ff GG HH 2(gg) + ff GG NN 2(gg) rr GG = 2 6.4 3 0 kk + 0 rr GG = 32.8
(b) Step 4: Calculate the equilibrium constant from the standard free energy change rr GG = RRRRRRRRRR llllll = rrgg RRRR llllll = 32.8 000 8.34 462 KK (298.5 KK) llllll = 3.2 KK = ee 3.2 KK = 6 0 5 Step 5: Check your work Does your answer seem reasonable? This is another two temperatures; two equilibrium constants question requiring the use of. Again, keeping organized is key to success. Some of the steps were KK RR TT TT 2 already taken care of in part (a). Step : Match equilibrium constants to the corresponding temperature TT = 298.5 KK TT 2 = 300 = 573.5 KK KK = 6 0 5 KK 2 =??? Step 2: Calculate the standard enthalpy change for the reaction rr HH = ff HH (pppppppppppppppp) ff HH (rrrrrrrrrrrrrrrrrr) rr HH = 2 ff HH NNNN 3(gg) 3 ff HH HH 2(gg) + ff HH NN 2(gg) rr HH = 2 45.9 3 0 + 0 rr HH = 9.8 = 9.8 04 Step 3: Crunch the numbers KK RR TT TT 2 6 0 5 = 9.8 04 298.5 KK 8.34 462 KK 573.5 KK 6 05 = 7.8 KK 2 = 6 0 5 ee 7.8 KK 2 = 2 6 0 5 0 8 KK 2 = (6 0 5 )(2 0 8 ) KK 2 = 0 2 Step 4: Check your work Does your answer seem reasonable? This reaction is exothermic ( rr HH < 0), so it is expected to have a smaller equilibrium constant at the higher temperature.
(c) (d) (e) The formula used to calculate K2 was derived assuming that the enthalpy change and entropy change for a reaction are constant at all temperatures. This is a reasonable approximation but is not actually true. Since the value for rh varies somewhat with temperature, this formula can only be used to calculate approximate equilibrium constants. The equilibrium constant for this reaction decreases as the temperature of reaction increases. This means that, at a lower temperature, more product can form before equilibrium is reached. Since you have already calculated K at 300, you can determine whether or not the reaction is thermodynamically allowed at that temperature by calculating Q and comparing it to K. Step : Determine activities for all reactants and products 50 bbbbbb aa NN2(gg) = = 50 aa 30 bbbbbb bbbbbb NNNN 3(gg) = = 30 bbbbbb 50 bbbbbb aa HH2(gg) = = 50 bbbbbb Step 2: Calculate reaction quotient QQ = aa 2 NNNN 3(gg) = (30)2 = 5.3 aa NN2(gg) aa HH2(gg) 3 0 6 (50)(50) 3 Step 3: Compare Q to K (at the appropriate temperature) QQ = 5.3 0 6 KK = 0 2 QQ < KK therefore the reaction is thermodynamically allowed under these conditions.
8. (a) (b) sublimation This pressure is BELOW the triple point, so liquid phase is not possible!
(b) sublimation This pressure is BELOW the triple point, so liquid phase is not possible!
9. (a) (b) (c) FALSE Solutes increase the boiling point of solutions compared to the pure solvent. Since they decrease the vapor pressure of the solvent at any given temperature, it requires a higher temperature for the vapor pressure to be equal to the atmospheric pressure (which is when the liquid boils). FALSE The free energy of the system decreases in a thermodynamically allowed process. The entropy of the universe increases in a thermodynamically allowed process, and rr GG = TT SS uuuuuuuuuuuuuuuu FALSE At equilibrium, Q = K. Under standard conditions, Q = (since all activities are under standard conditions). 0. (a) nonideal gas ideal gas Recall from CHEM 000 that a gas behaves ideally when it has low density (therefore its particles take up negligible volume and experience negligible intermolecular forces). This occurs at relatively low pressure and or/high temperature. Nonideal behavior is observed at pressures almost (but not quite) high enough to compress the gas to a liquid. (b) A supercritical fluid is a substance that is at a high enough pressure that it has the density of a liquid and a high enough temperature that its particles have the high average kinetic energy that would usually be associated with a gas. (c) i. see the green dot on the diagram above ii. At 0 bar and 300 K, CO2 is a gas. As the temperature is decreased, it condenses to a liquid then freezes to a solid. iii. As the temperature decreases, the average kinetic energy of particles in the sample decreases. Having less kinetic energy to overcome intermolecular forces, the particles in the sample pack more densely so a liquid (then, at lower temperatures, a solid) forms.
. (a) (b) (c) Vapour pressure is the pressure exerted by molecules which have evaporated (or sublimed) out of a liquid (or solid) sample. Vapour pressure increases as temperature increases. As temperature increases, the average kinetic energy of the particles in the sample increases. Therefore, more particles within the sample will have enough energy to overcome intermolecular forces within the liquid (or solid) and escape into the gas phase. As the number of particles in the gas phase increases, the pressure they exert increases. Step : Write out the relevant balanced chemical equation (the reaction is liquid X evaporates to give gas X) XX (ll) XX (gg) Step 2: Write the equilibrium constant expression for this reaction KK = aa XX (gg) aa XX(ll) Step 3: Fill in the activities aa XX(ll) = KK = aa XX (gg) aa XX(ll) = pp XX(gg) bbbbbb = pp XX (gg) bbbbbb aa XX(gg) = pp XX (gg) bbbbbb Step 4: Substitute for K at each temperature and simplify to give the Clausius-Clapeyron equation. (p is the vapour pressure at T ; K is the equilibrium constant at T) KK RR TT TT 2 llnn pp2 bbbbbb pp bbbbbb RR TT TT 2 llll pp 2 pp RR TT TT 2 2. (a) (b) KK = aa 2 NNNN (gg) aacccc2(gg) aa NNNNNNNN(gg) 2 Equilibrium constants can be calculated from the standard free energy change for a reaction. When this question was originally on a test, standard free energies of formation were not provided. With that limited set of data, it was best to calculate the standard enthalpy change for the reaction and the standard entropy change for the reaction then use those values to calculate the standard free energy change for the reaction. The standard enthalpy change for the reaction was necessary for part (c) anyway. Step : Write a balanced chemical equation for the reaction This was provided in the question. 2 NNNNNNNN (gg) 2 NNNN (gg) + CCCC 2(gg)
Step 2: Convert temperature to Kelvin (if necessary) TT = 25 = 298.5 KK Step 3: Calculate the standard free energy change for the reaction rr GG = ff GG (pppppppppppppppp) ff GG (rrrrrrrrrrrrrrrrrr) (c) rr GG = 2 ff GG NNNN (gg) + ff GG CCCC 2(gg) 2 ff GG NNNNNNNN (gg) rr GG = 2 86.55 + 0 2 66.08 rr GG = 40.94 Step 4: Calculate the equilibrium constant from the standard free energy change rr GG = RRRRRRRRRR llllll = rrgg RRRR llllll = 40.94 000 8.34 462 KK (298.5 KK) llllll = 6.52 KK = ee 6.52 KK = 6.7 0 8 Step 5: Check your work Does your answer seem reasonable? A positive standard free energy change should correspond to a small equilibrium constant (i.e. significantly smaller than ). This is another two temperatures; two equilibrium constants question requiring the use of. Again, keeping organized is key to success. KK RR TT TT 2 Step : Match equilibrium constants to the corresponding temperature TT = 298.5 KK TT 2 = 745 KK KK = 6.7 0 8 KK 2 =??? Step 2: Calculate the standard enthalpy change for the reaction rr HH = ff HH (pppppppppppppppp) ff HH (rrrrrrrrrrrrrrrrrr) rr HH = 2 ff HH NNNN (gg) + ff HH CCCC 2(gg) 2 ff HH NNOOOOOO (gg) rr HH = 2 90.25 + 0 2 5.7 rr HH = +77.08 = +7.708 04 Step 3: Crunch the numbers KK RR TT TT 2 KK llll 2 6.7 0 8 = 7.708 04 8.34 462 KK KK llll 2 6.7 0 8 = 8.6 KK 2 = ee8.6 6.7 0 8 298.5 KK 745 KK
KK 2 6.7 0 8 = 08 KK 2 = (6.7 0 8 )( 0 8 ) KK 2 = 8 Step 4: Check your work Does your answer seem reasonable? This reaction is endothermic ( rr HH > 0), so it is expected to have a larger equilibrium constant at the higher temperature. Alternatively, the question could be solved by calculating the free energy change at 745 K from the standard enthalpy change and standard entropy change (assuming that those two values do not vary significantly with temperature) using GG = HH TT SS. The free energy change at 745 K could then be used to calculate the equilibrium constant at that temperature using GG = RRRRRRRRRR. (This formula can be used at other temperatures in which case, all activities must still be one.) Please note that you CANNOT just use the standard free energy change calculated for 298.5 K in the above equation with T = 745. Free energy varies significantly with temperature! The temperature at which standard free energy is calculated must always match the temperature used in GG = RRRRRRRRRR for the formula to be applicable. This alternative approach also gives KK 2 = 8. (d) Since you have already calculated K at 745 KK, you can determine whether or not the reaction is thermodynamically allowed at that temperature by calculating Q and comparing it to K. Step : Determine activities for all reactants and products 0.05 bbbbbb 0.4 bbbbbb aa NNNNNNNN(gg) = = 0.05 aa NNNN(gg) = = 0.4 bbbbbb Step 2: Calculate reaction quotient aa CCCC2(gg) = bbbbbb 0.32 bbbbbb bbbbbb = 0.32 QQ = aa 2 NNNN (gg) aacccc2(gg) aa NNNNNNNN(gg) 2 = (0.4)2(0.32) = 22 = 2 0 (0.05) 2 Step 3: Compare Q to K (at the appropriate temperature) QQ = 2 0 KK = 8 QQ > KK therefore the reaction is not thermodynamically allowed in the forward direction under these conditions. Instead, it will proceed in reverse (assuming that the kinetics allow for that).
3. (a) (b) 89.3 kk By definition, the vapour pressure at a boiling point is equal to the atmospheric pressure. This question is most easily answered using Raoult s Law. Begin by calculating the mole fraction of water. Step : Calculate the moles of each species in solution nn HH2 OO = 000 gg = 55.5 HH 8.052 gg 2OO To calculate the moles of each ion in solution, it is easiest to calculate the moles of MMMMMMMM 2 then use mole ratios to calculate the moles of MMMM 2+ and CCCC. nn MMMM 2+ = 0.50 MMMMMMMM 2 nn CCll = 0.50 MMMMMMMM 2 MMMM2+ MMMMMMMM 2 2 CCCC MMMMMMMM 2 = 0.50 MMMM2+ =.0 CCCC Step 2: Calculate the mole fraction of water in the solution XX HH2 OO = nn HH 2OO nn HH 2OO + nn MMMM 2+ +(nn CCCC ) = 55.5 (55.5 )+(0.50 )+(.0 ) = 0.974 Step 3: Calculate the vapour pressure above the solution pp HH2 OO = XX HH2 OO pp HH2 OO = (0.974)(89.3 kk) = 87.0 kk Step 4: Check your work Does your answer seem reasonable? Are sig. fig. correct? The dissolved salt lowers the vapour pressure of water above the solution. 4. Equilibrium constants can be calculated from the standard free energy change for a reaction. Step : Write a balanced chemical equation for the reaction NN 2 OO 4(gg) 2 NNNN 2(gg) Step 2: Convert temperature to Kelvin (if necessary) TT = 25 = 298.5 KK Step 3: Calculate the standard free energy change for the reaction rr GG = ff GG (pppppppppppppppp) ff GG (rrrrrrrrrrrrrrrrrr) rr GG = 2 ff GG NNNN 2(gg) ff GG NN 2 OO 4(gg) rr GG = 2 5.3 97.89 rr GG = 4.73 Step 4: Calculate the equilibrium constant from the standard free energy change rr GG = RRRRRRRRRR llllll = rrgg RRRR llllll = llllll =.9 4.73 000 8.34 462 KK (298.5 KK)
KK = ee.9 KK = 0.5 Step 5: Check your work Does your answer seem reasonable? A positive standard free energy change should correspond to a small equilibrium constant (i.e. smaller than ). Now, the equilibrium constant at 25 can be used to calculate the temperature corresponding to a different equilibrium constant. That makes this another two temperatures; two equilibrium constants question requiring the use of. KK RR TT TT 2 Start by calculating the equilibrium constant of interest (i.e. the one corresponding to the two pressures described in the question). Step 6: Determine activities for all reactants and products 0.5 bbbbbb aa NN2 OO 4(gg) = = 0.5 aa 0.5 bbbbbb bbbbbb NNNN 2(gg) = = 0.5 bbbbbb Step 7: Calculate the equilibrium constant for the pressures provided. This will be K2. (The equilibrium constant at 2222 is K.) KK 2 = aa 2 NNNN 2(gg) aa NN 2OO 4(gg) KK 2 = (0.5)2 = 0.5 (0.5) Step 8: Match equilibrium constants to the corresponding temperature TT = 298.5 KK TT 2 =??? KK = 0.5 KK 2 = 0.5 Step 9: Calculate the standard enthalpy change for the reaction rr HH = ff HH (pppppppppppppppp) ff HH (rrrrrrrrrrrrrrrrrr) rr HH = 2 ff HH NNNN 2(gg) ff HH NN 2 OO 4(gg) rr HH = 2 33.8 9.6 rr HH = + 57.20 = +5.720 04 Step 0: Crunch the numbers KK RR TT TT 2 llll 0.5 = 5.720 04 0.5 298.5 KK TT 2 8.34 462 KK llll 0.5 0.5 8.34 462 KK 5.720 0 4 = TT 2 298.5 KK llll = 298.5 KK TT 2 0.5 0.5 8.34 462 KK 5.720 0 4 = 0.003 354 0 0.000 8 TT 2 KK KK = 0.003 8 TT 2 KK
TT 2 = 0.003 8 KK = 35 KK = 42 Step : Check your work Does your answer seem reasonable? The desired equilibrium constant is relatively close to the equilibrium constant at 25, so it is reasonable that it will be achieved at a temperature close to 25. 5. (a) Equilibrium constants can be calculated from the standard free energy change for a reaction. Vapour pressure can be calculated from the equilibrium constant for evaporation. Step : Write a balanced chemical equation for the reaction (or phase change) CCCC 2(ll) CCCC 2(gg) Step 2: Convert temperature to Kelvin (if necessary) TT = 25 = 298.5 KK Step 3: Calculate the standard free energy change for the reaction (or phase change) rr GG = ff GG (pppppppppppppppp) ff GG (rrrrrrrrrrrrrrrrrr) rr GG = ff GG CCSS 2(gg) ff GG CCCC 2(ll) rr GG = 67. 64.6 rr GG = 2.5 Step 4: Calculate the equilibrium constant from the standard free energy change rr GG = RRRRRRRRRR llllll = rrgg RRRR llllll = 2.5 000 8.34 462 KK (298.5 KK) llllll =.0 KK = ee.0 KK = 0.4 Step 5: Use equilibrium constant expression to calculate vapour pressure KK = aa CCCC 2(gg) aa CCCC2(ll) pp CCCC2(gg) bbbbbb KK = pp CCCC2(gg) = KK bbbbbb = 0.4 bbbbbb Step 6: Check your work Does your answer seem reasonable? A positive standard free energy change should correspond to a small equilibrium constant (i.e. smaller than ). Since CS2 is a liquid at room temperature, its vapour pressure should be lower than atmospheric pressure at room temperature.
(b) Normal boiling point is the temperature at which vapour pressure is atm. That knowledge can be used to calculate the equilibrium constant at the boiling point. From there, this becomes another two temperatures; two equilibrium constants question requiring the use of. KK RR TT TT 2 Step : Determine activities for all reactants and products aa CCCC2(ll) = aa CCCC2(gg) = aaaaaa.0325 bbbbbb =.0325 bbbbbb aaaaaa Step 2: Calculate the equilibrium constant for the pressures provided. This will be K2. (The equilibrium constant at 2222 is K. This was calculated in part (a).) KK 2 = aa CCCC 2(gg) aa CCCC2(ll) KK 2 = (.0325) =.0325 () Step 3: Match equilibrium constants to the corresponding temperature TT = 298.5 KK TT 2 =??? KK = 0.36 KK 2 =.0325 Step 4: Calculate the standard enthalpy change for the reaction rr HH = ff HH (pppppppppppppppp) ff HH (rrrrrrrrrrrrrrrrrr) rr HH = ff HH CCCC 2(gg) ff HH CCCC 2(ll) rr HH = 6.7 89.0 rr HH = +27.7 = +2.77 04 Step 5: Crunch the numbers KK RR TT TT 2 llll.0325 = 2.77 04 0.36 298.5 KK TT 2 8.34 462 KK llll.0325 0.36 8.34 462 KK 2.77 0 4 = llll TT 2 298.5 KK = 298.5 KK TT 2.0325 0.36 8.34 462 2.77 0 4 KK = 0.003 354 0 0.000 307 TT 2 KK KK = 0.003 047 TT 2 KK TT 2 = 0.003 047 KK = 328.2 KK = 55.0 Step 6: Check your work Does your answer seem reasonable? The desired equilibrium constant is relatively close to the equilibrium constant at 25, so it is reasonable that it will be achieved at a temperature close to 25. Since we are told that CS2 is a liquid at room temperature, its boiling point must be higher than room temperature.
6. (a) (b) Equilibrium constants can be calculated from the standard free energy change for a reaction. Vapour pressure can be calculated from the equilibrium constant for evaporation. Step : Write a balanced chemical equation for the reaction (or phase change) CCCCCC 4(ll) CCCCCC 4(gg) Step 2: Convert temperature to Kelvin (if necessary) TT = 25 = 298.5 KK Step 3: Calculate the standard free energy change for the reaction (or phase change) rr GG = ff GG (pppppppppppppppp) ff GG (rrrrrrrrrrrrrrrrrr) rr GG = ff GG CCCCCC 4(gg) ff GG CCCCCC 4(ll) rr GG = 60.59 65.2 rr GG = 4.62 Step 4: Calculate the equilibrium constant from the standard free energy change rr GG = RRRRRRRRRR llllll = rrgg RRRR llllll = 4.62 000 8.34 462 KK (298.5 KK) llllll =.86 KK = ee.86 KK = 0.6 Step 5: Use equilibrium constant expression to calculate vapour pressure KK = aa CCCCCC 4(gg) aa CCCCCC4(ll) pp CCCCCC4(gg) bbbbbb KK = pp CCCC2(gg) = KK bbbbbb = 0.6 bbbbbb Step 6: Check your work Does your answer seem reasonable? A positive standard free energy change should correspond to a small equilibrium constant (i.e. smaller than ). Liquid Since the vapour pressure of CCCCCC 4 is lower than atmospheric pressure at 25, this temperature must be lower than the boiling point of CCCCCC 4. Alternatively, some students argued in favour of the liquid phase based on the calculated rr GG. They stated that, because it was a positive value, the conversion of liquid to gas phase was not favoured at 25.
(c) This question is most easily answered using Raoult s Law. Begin by calculating the mole fraction of CCCCCC 4. Step : Calculate the moles of each species in solution nn CCCCCC4 = 250 gg =.63 CCCCCC 53.822 gg 4 nn CC6 HH 6 = 20.0 gg = 0.256 CC 78.3 gg 6HH 6 Step 2: Calculate the mole fraction of water in the solution XX CCCCCC4 = nn CCCCCC 4 nn CCCCCC 4 + nn CC6HH6 =.63 (.63 )+(0.256 ) = 0.864 Step 3: Calculate the vapour pressure above the solution pp HH2 OO = XX HH2 OO pp HH2 OO = (0.864)(0.6 bbbbbb) = 0.3 bbbbbb Step 4: Check your work Does your answer seem reasonable? Are sig. fig. correct? The dissolved benzene lowers the vapour pressure of CCCCCC 4 above the solution. 7. This question is most easily answered using Henry s Law. Use the data in the first sentence to calculate the Henry s Law constant (kk HH ) for carbon dioxide. Use that constant to answer the question in the second sentence. Step : Write a balanced chemical equation for the reaction (or phase change) CCCC 2(gg) CCCC 2(aaaa) Step 2: Calculate the Henry s Law constant for carbon dioxide [CCCC 2 ] = kk HH pp CCCC2 kk HH = [CCCC 2 ] pp CCCC 2 = 0.035 LL = 0.034 (.03 aaaaaa) LL aaaaaa Step 3: Calculate solubility of carbon dioxide with a different partial pressure [CCCC 2 ] = kk HH pp CCCC2 [CCCC 2 ] = 0.034 LL aaaaaa (2.24 aaaaaa) [CCCC 2 ] = 0.0763 LL Step 4: Check your work Does your answer seem reasonable? Are sig. fig. correct? A higher partial pressure of carbon dioxide gas over the solution ought to result in a higher concentration of dissolved carbon dioxide. Since the pressure is approximately doubled, the concentration ought to be approximately doubled too.
8. (a) (b) Equilibrium constants can be calculated from the standard free energy change for a reaction. KK aa is the equilibrium constant for dissociation of HH + from an acid. pppp aa is readily calculated from KK aa. Step : Write a balanced chemical equation for the reaction (or phase change) + HHHHHH (aaaa) HH (aaaa) + CCCC (aaaa) Step 2: Convert temperature to Kelvin (if necessary) TT = 25.00 = 298.5 KK Step 3: Calculate the standard free energy change for the reaction (or phase change) rr GG = ff GG (pppppppppppppppp) ff GG (rrrrrrrrrrrrrrrrrr) rr GG = ff GG HH + (aaaa) + ff GG CCCC (aaaa) ff GG HHHHHH (aaaa) rr GG = 0 + 66 2 rr GG = 54 Step 4: Calculate the equilibrium constant from the standard free energy change rr GG = RRRRRRRRRR llllll = rrgg RRRR llllll = 54 000 8.34 462 KK (298.5 KK) llllll = 22 KK = ee 22 KK = 3 0 0 Step 5: Calculate pppp aa from KK aa pppp aa = llllll(kk aa ) = llllll(3 0 0 ) = 9.5 Step 6: Check your work Does your answer seem reasonable? A positive standard free energy change should correspond to a small equilibrium constant (i.e. smaller than ). HCN is a weak acid, so it should have a pppp aa value greater than 0. (Recall definition of strong vs. weak acid from CHEM 000.) This is another two temperatures; two equilibrium constants question requiring the use of. As always, keeping organized is key to success. KK RR TT TT 2 Step : Match equilibrium constants to the corresponding temperature TT = 298.5 KK TT 2 = 323.5 KK KK = 3 0 0 KK 2 =??? Step 2: Calculate the standard enthalpy change for the reaction rr HH = ff HH (pppppppppppppppp) ff HH (rrrrrrrrrrrrrrrrrr) rr HH = ff HH HH + (aaaa) + ff HH CCCC (aaaa) ff HH HHHHHH (aaaa) rr HH = 0 + 5 05 rr HH = +46 = +4.6 04
Step 3: Crunch the numbers KK RR TT TT 2 KK llll 2 3 0 0 = 4.6 04 KK llll 2 3 0 KK 2 3 0 0 = ee.4 0 =.4 8.34 462 KK 298.5 KK 323.5 KK KK 2 3 0 0 = 4 KK 2 = (3 0 0 )(4) KK 2 = 0 9 Step 4: Check your work Does your answer seem reasonable? This reaction is endothermic ( rr HH > 0), so it is expected to have a larger equilibrium constant at the higher temperature. Step 5: Answer the question! There was a verbal question attached to this calculation. It s easy to forget about that. Always re-read the question when you think you re done to make sure you answered everything! HCN is a stronger acid at the higher temperature. The higher equilibrium constant means that it dissociates to a greater degree at 50.00.
9. (a) (b) Equilibrium constants can be calculated from the standard free energy change for a reaction. KK aa is the equilibrium constant for dissociation of HH + from an acid. Step : Write a balanced chemical equation for the reaction (or phase change) + HH 2 CCCC 3(aaaa) HH (aaaa) + HHHHHH 3(aaaa) Step 2: Convert temperature to Kelvin (if necessary) TT = 25.00 = 298.5 KK Step 3: Calculate the standard free energy change for the reaction (or phase change) rr GG = ff GG (pppppppppppppppp) ff GG (rrrrrrrrrrrrrrrrrr) rr GG = ff GG HH + (aaaa) + ff GG HHHHHH 3(aaaa) ff GG HH 2 CCCC 3(aaaa) rr GG = 0 + 586.8 623. mmooll rr GG = 36.3 Step 4: Calculate the equilibrium constant from the standard free energy change rr GG = RRRRRRRRRR llllll = rrgg RRRR llllll = 36.3 000 8.34 462 KK (298.5 KK) llllll = 4.6 KK = ee 4.6 KK = 4 0 7 Step 5: Check your work Does your answer seem reasonable? A positive standard free energy change should correspond to a small equilibrium constant (i.e. smaller than ). HH 2 CCCC 3 is a weak acid, so it should have a KK aa value less than. rr HH = ff HH (pppppppppppppppp) ff HH (rrrrrrrrrrrrrrrrrr) rr HH = ff HH HH + (aaaa) + ff HH HHHHHH 3(aaaa) ff HH HH 2 CCCC 3(aaaa) rr HH = 0 + 689.9 699.7 rr HH = +9.8 Since rr HH > 0, this reaction is endothermic, so increasing the temperature will favour the forward reaction. This corresponds to an increase in the equilibrium constant (KK aa ) and a strengthening of the acid (since more H + is present at equilibrium).