Chapter 6 Answers Page 195 #1. Use the tables on page 420 (heat of dissolution) and 422 (polyatomic ions) to help you answer these questions. a) HClO3(s) H+(aq) +ClO3-(aq) ΔH= b) HI(s) H+(aq) + I-(aq) ΔH= c) KNO3(s)K+(aq) +NO3-(aq) ΔH= d) CuSO4(s) Cu2+(aq) + SO42-(aq) ΔH= e) Li2CO3(s) 2 Li+(aq) + CO3 2-(aq) ΔH= -42.0 -30.0 +36.0 -68.0 -13.0 exothermic exothermic endothermic exothermic exothermic Page 195, #2 Step 1: From the problem we identify that the mass of water is 20g, the ΔT is 50°C and the c is 4.184 DATA Step 2: Find Q Q=mwcw ΔT Calorimeter: =20 g x4.184 J/g°C x 50°C mw=20g =4184.0 J cw=4.184 J/g°C Step 3: Calculate the number of moles of the solute ΔT=50°C n=ms / Ms =4g / 39.99g/mol Solute: =0.100 mol mS = 4g Step 4: Convert Joules to kiloJoules and determine sign MS= 39.99g/mol 4184.0 J = 4.184 kJ The reaction is exothermic so the sign is negative To find: n= Q= ΔH= Step 5: Calculate ΔH Answer: ΔH = -41.8 kJ/mol ΔH= Q / n = -4.184 kJ / 0.100 mol = -41.8 kJ/mol Page 195 #3 Step 1: calculate heat... Q = mwcwΔT mw=1000g = 1000g x 4.184 J/g°C x -2°C Cw=4.184 = -8368 J ΔT=-2°C Step 2: Determine the sign... And change to kJ ms=24.42g endothermic, so positive +8.368 kJ Ms=122.5 Step 3: Find the number of moles of KClO3 ns = ms / Ms find = 24.42g / 122.5 g.mol-1 ns= = 0.1993 mol Step 4: Find ΔHd = Q / ns = 8.368 kJ / 0.199 mol =41.9 kJ/mol Data Page 195 #4 • Dissolution takes place in two steps. In the first step, solute particles are separated from each other. This process is endothermic (absorbs energy from the water, positive enthalpy change). The second step the solute particles are distributed through the solvent. This step is exothermic (releases energy, negative enthalpy change). Page 196 #5 Data mw=1000g Cw=4.184 ΔT=5.5°C n= 1 mol Step 1: Calculate heat... Q=mwcwΔT = 1000g x 4.184 J/g°C x 5.5°C Q= = 23 012 J or 23.012 kJ Step 2: Determine the sign... It’s positive Step 3: find the ΔHd... ΔHd = Q / n (but n=1, so that’s easy) ΔHd = +23.012 Step 4: look it up on page 420... The only salt with a ΔHd of about +23 is AgNO3, (silver nitrate) Data mw=100g Cw=4.184 Ti=20°C ms=8g. Ms=159.6 ΔHd=68.0 Find ns= ΔT= Tf = Page 196 #6 Step 1: find the number of moles of CuSO4 n= ms / Ms = 8g / 159.6 = 0.0501 mol Step 2: calculate the heat using the table p. 420 Q= n x ΔHd = 0.0501 mol x 68 kJ/mol = 3.400 kJ or 3400 J Step 3: calculate ΔT = Q / mwcw = 3400 J / (100g•4.18 J/g°C) = 8.13°C Step 4: Calculate Tf = 20 + 8.13 = 28.13°C Round answer to a sensible number Answer: final temperature is 28°C Data mw=200g Cw=4.184 Ti=22 ms=8.42g. Ms= 42.4 ΔHd= Find ns= ΔT= Tf = Page 196 #7 Step 1: Find number of moles of solute ns=ms / Ms = 8.42g / 42.4 g/mol =0.1986 mol Step 2: Calculate the heat from dissolution Q = ΔHd x ns = -35 kJ/mol x 0.1986 mol = -6.951 or -6951 J Step 3: Determine effect (Exothermic = temperature up) Step 4: Calculate temperature change, Q=mwcwΔT, so... ΔT= Q / mc = 6951J / (200•4.184) = 8.31°C Step 5: Calculate Tf... = 22° + 8.31 ° =30.31°C Page 196, #8, 8. Step 1: Step 2: ΔT Q = 80°C – 25°C = 55°C = mcΔT = 100 (4.184)(55) = 23 012 J or 23.012 kJ Step 3: its exothermic, so it will be negative Step 4: n = Q / ΔH, = -23.012 / -55.0 = 0.4184 mol Step 5: m = nM = 0.4184 x 56.1 =23.47 g The necessary amount is 23g of NaOH Page 195, #9 Steps: 1. Calculate ΔT ΔT = Tf – Ti 2. Calculate energy Q=mc ΔT 3. Determine sign endothermic 4. Calculate moles, if the salt were NaCl n = m/M = -7°C =-1464.4 J ΔHd+ = 10/58.4 = 0.1712 mol 5. Calculate ΔHd ΔHd= Q/n = 4.3 Since ΔHd of NaCl is +4.3 kJ/mol, and the salt used absorbs 8.56 kJ/mol, it is not the same salt Page 195, #10 Steps: 1. Find number of moles n= C(mol/L)V(L) 2. Calculate heat transfer Q=n ΔHd 3. Determine effect endothermic 4. Calculate ΔT ΔT = Q / mc 5. Find final temperature The final temperature is 23.5°C More Answers (195, #11-13) 11.Answer: The molar heat of neutralization is: -0.6 kJ/mol 12.Answer: 90g. Of sodium must be dissolved 13.Answer: The molar heat of neutralization is: -0.6 kJ/mol 14.Answer: 57g of sodium hydroxide is required to heat the meal.