Answer Key ALE 28. ess s Law and Standard Enthalpies Frmatin (Reerence: Chapter 6 - Silberberg 4 th editin) Imprtant!! Fr answers that invlve a calculatin yu must shw yur wrk neatly using dimensinal analysis with crrect signiicant igures and units t receive ull credit. N wrk, n credit. Reprt numerical answers t the crrect number signiicant igures. CIRCLE ALL NUMERICAL RESPONSES. w can standard heats rmatin be used t calculate the heat a reactin? The Mdel: Algebraic Additin Standard eats Frmatin via ess s Law In the last ALE, we lked at hw ess s law culd be used t add thermchemical equatins tgether t determine the heat a reactin interest. We saw that smetimes it is apprpriate t add atmizatin reactins tgether and at ther times it is apprpriate t add cmbustin reactins tgether. Since enthalpy is a state unctin, it makes n dierence what kind thermchemical equatins are added tgether. The r a speciic chemical reactin is a speciic value that is determined by the enthalpies prducts and reactants. Anther example, this time using thermchemical equatins rmatin reactins, is the reactin between ethyne (als knwn as acetylene) and nitrgen mnxide: C 2 2 (g) + 5 NO(g) 2 CO 2 (g) + 5 / 2 N 2 (g) + 2 O(g) (Eqn 1) The bank enthalpies rmatin reactins are: 2 C(s, graphite) + 2 (g) C 2 2 (g) (C 2 2 (g)) = 226.8 kj ½ N 2 (g) + ½ O 2 (g) NO(g) (NO(g)) = 90.3 kj C(s, graphite) + O 2 (g) CO 2 (g) (CO 2 (g)) = -393.5 kj 2 (g) + ½ O 2 (g) 2 O(g) ( 2 O(g)) = -238.9 kj The equatins are added tgether in the llwing way: C 2 2 (g) 2 C(s, graphite) + 2 (g) - (C 2 2 (g)) = -226.8 kj 5 NO(g) 5 / 2 N 2 (g) + 5 / 2 O 2 (g) -5 (NO(g)) = -451.5 kj 2 C(s, graphite) + 2 O 2 (g) 2 CO 2 (g) 2 (CO 2 (g)) = -787.0 kj 2 (g) + ½ O 2 (g) 2 O(g) ( 2 O(g)) = -238.9 kj = [-226.8 + (-451.5) + (-787.0) + (-238.9)] kj = -1704.2 kj And the standard-state heat,, reactin 1 is calculated t be -1704.2 kj per mle reacted ethyne. Key Questins 1. When the reactins are added tgether, the reactins invlving the reactants are (circle the answer): i. written as rmatin reactins. ii. written as decmpsitin reactins. 2. When the reactins are added tgether, the reactins invlving the prducts are (circle the answer): i. written as rmatin reactins. ii. written as decmpsitin reactins. Page 1 5
3. When the reactin heats are added tgether, the heat reactin invlving a reactant (circle the answer): i. maintains the same sign as the heat rmatin r the substance. ii. changes sign rm what it had as a heat rmatin. 4. When the reactin heats are added tgether, the heat reactin invlving a prduct (circle the answer): i. maintains the same sign as the heat rmatin r the substance. ii. changes sign rm what it had as a heat rmatin. 5. What happens t r a bank reactin when the reactin is multiplied thrugh by a cnstant? When a reactin is multiplied thrugh by a cnstant, that reactin als gets mulitplied by the same cnstant. ( Flipping a reactin may be thught as mulitplying all stichimetric ceicients by -1, s changes sign.) The Mdel: Anther Way Applying ess s Law Fr any balanced chemical equatin, the heat the reactin can be calculated by the rmula = n p (prducts) n r (reactants) where n p and n r are the stichimetric ceicients prducts and reactants, respectively, and is the heat rmatin a prduct r reactant. (The cllectin symbls n (X(g)) is read n times the standard heat rmatin gaseus X.) The standard heats rmatin the prducts r reactants can be lked up in a table thermdynamic data, such Appendix B in yur textbk. Key Questins 6. ess s Law allws the heat reactin 1 (i.e., the reactin between ethyne and nitrgen mnxide) t be calculated thrugh the use the rmula: sum ver prducts = 2 (CO 2 (g)) + 5 / 2 (N 2 (g)) + ( 2 O(g)) (C 2 2 (g)) 5 (NO(g)) a. Explain why the heats rmatin CO 2, N 2, and 2 O are added while the heats rmatin C 2 2 and NO are subtracted. (int: Revisit Questins 3 and 4.) sum ver reactants CO 2, N 2, and 2 O are the prducts, and C 2 2 and NO are the reactants. The heats rmatin the reactants are subtracted rm the heats rmatin the prducts (which are added tgether). ALE 28 - Page 2 5
6b. Explain the ceicients bere each. (i.e. Why is the heat rmatin CO 2 multiplied by 2? Why is the heat rmatin NO multiplied by 5? int: Revisit Questin 5.) The stichimetric ceicient in rnt C 2 2 is -1 (ethyne is a reactant), s the heat rmatin C 2 2 is multiplied by -1. The stichimetric ceicient in rnt NO is -5 (nitrgen mnxide is a reactant), s the heat rmatin NO is multiplied by -5. The stichimetric ceicient in rnt CO 2 is 2, s the heat rmatin CO 2 is multiplied by 2. The stichimetric ceicient in rnt N 2 is 5 / 2, s the heat rmatin N 2 is multiplied by 5 / 2. And since the stichimetric ceicient 2 O is 1, it heat rmatin is just itsel. Exercises A. Use the table thermdynamic data in Appendix B yur textbk t calculate the standard heat reactin r the llwing reactins at 25 C. Shw yur wrk and circle yur answer. 2 N 3 (g) + 3 NO(g) 5 / 2 N 2 (g) + 3 2 O(g) = 5 / 2 (N 2 (g)) + 3 ( 2 O(g)) 2 (N 3 (g)) 3 (NO(g)) = [( 5 / 2 )(0) + (3)(-241.8) (2)(-46.3) (3)(90.4)] kj/ml = -904.0 kj/ml NaCO 3 (s) Na 2 CO 3 (s) + 2 O(l) + CO 2 (g) (UNBALANCED!) Balanced: 2 NaCO 3 (s) Na 2 CO 3 (s) + 2 O(l) + CO 2 (g) = (Na 2 CO 3 (s)) + ( 2 O(l)) + (CO 2 (g)) 2 (NaCO 3 (s)) = [(-1130.9) + (-285.8) + (-393.5) (2)(-947.68)] kj/ml = +85.2 kj/ml MgCl 2 (aq) + 2 NaO(aq) Mg(O) 2 (s) + 2 NaCl(aq) ints: Write the net inic equatin, paying attentin t the phase the reactants and prducts. Use the net inic equatin t calculate. The Mg(O) 2 (s) is -924.66 kj/ml. Net Inic: Mg 2+ (aq) + 2 O - (aq) Mg(O) 2 (s) = (Mg(O) 2 (s)) (Mg 2+ (aq)) 2 (O - (aq)) = [(-924.66) (-461.96) (2)(-229.94)] kj/ml = -2.82 kj/ml ALE 28 - Page 3 5
B. An instant ice pack when purchased has ne cmpartment illed with slid ammnium nitrate and anther cmpartment illed with water. When a seal between the tw cmpartments is brken, the water and ammnium nitrate mix, disslving the salt: N 4 NO 3 (s) 2 O N 4 + (aq) + NO 3 - (aq) Calculate (in kj/ml) r the slvatin ammnium nitrate. ints: slid ammnium nitrate at 25 C is -365.56 kj/ml, the aqueus nitrate at 25 C is -205.0 kj/ml and the aqueus ammnium at 25 C is -133.8 kj/ml. 2 O is written abve the reactin arrw t let yu knw it is necessary r the slvatin t ccur, but yu shuld nt cnsider water t be a reactant. Shw yur wrk and circle yur answer. = (N + 4 (aq)) + (NO - 3 (aq)) (N 4 NO 3 (s)) = [(-132.80) + (-205.0) (-365.56)] kj/ml = +27.8 kj/ml Suppse an instant ice pack has within it 36.2 g N 4 NO 3 (80.05 g/ml) and 254.6 ml water. (The density the water is 0.9978 g/ml and its mlar mass is 18.02 g/ml.) When the tw cmpnents mix, hw much heat is exchanged? Shw yur wrk and circle yur answer. n(n 4 NO 3 ) = (36.2 g)(1 ml / 80.05 g) = 0.452 ml q = n q = (0.452 ml N 4 NO 3 disslved)(27.8 kj absrbed / 1 ml N 4 NO 3 disslved) q = 12.6 kj absrbed Assume that n heat is exchanged between the slutin in the instant ice pack and its surrundings. (i.e. The reactin takes place in an insulated cntainer.) Suppse that the initial temperature the instant ice pack was 22.0 C. I the resultant aqueus slutin ammnium nitrate has a speciic heat 4.184 J/g C, what is the inal temperature the slutin? ints: Use the density water t cnvert the vlume water int the mass water. What is the mass the ammnium nitrate slutin? Frm the q calculated in and the knwn values m and C, calculate T. Is heat added t r taken rm the water? Des the temperature the water increase r decrease? Shw yur wrk and circle yur answer. Nte: sme chemists slve this prblem using the mass the slutin (as shwn belw) and thers will use nly the mass the water each methd having its advantages and disadvantages. m sln = m N4NO3 + m 2O = 36.2 g + (254.6 ml)(0.9978 g / 1 ml) = 290.2 g T = q / m sln c = (12.6 10 3 J) / (290.2 g)(4.18 J g -1 C -1 ) = 10.3 C Thermal energy lws rm the water int the ins in rder t separate them as they disslve, s T = T i T = 22.0 C 10.3 C = 11.7 C. ALE 28 - Page 4 5
C. Prblem 6.72: Make any changes needed in each the llwing equatins t make equal t, r the cmpund present in each equatin. (i.e. Rewrite the llwing equatins t make them rmatin equatins yu might want t revisit Key Questin #7 n ALE 27.) a.) Cl (g) + Na (s) NaCl (s) Frmatin equatin: ½ Cl 2 (g) + Na(s) NaCl(s) b.) 2 O (g) 2 (g) + ½ O 2(g) Frmatin equatin: 2 (g) + ½ O 2 (g) 2 O(l) c.) N 2(g) + 3 2(g) 2 N 3(g) Frmatin equatin: ½ N 2 (g) + 3 /2 2 (g) N 3 (g) D. Prblem 6.90: Physicians and nutritinal bichemists recmmend eating vegetable ils rather than animal ats t lwer risks heart disease. In live il, ne he healthier chices, the main atty acid is leic acid: C 18 34 O 2 ; cmb= -1.11 x 10 4 kj/ml. Use Appendix B and the cmb leic acid t calculate the leic acid. Start by writing the balanced chemical equatin r the cmbustin leic acid. Shw yur wrk and circle yur answer. Balanced chemical equatin: C 18 34 O 2 (l) + 51 /2 O 2 (g) 18 CO 2 (g) + 17 2 O(l) cmb= 1.11 x 10 4 kj/ml = 18 ml (, CO 2(g) ) + 17 ml (, 2 O (l) ) 51 /2 ml(, O 2(g) ) 1 ml (, leic acid) = 1.11 x 10 4 kj =18 ml (-393.5 kj/ml) + 17 ml (-285.840) acid) 51 /2 ml(0) 1 ml (, leic 1.11 x 10 4 kj = -11100 kj = 7083.0 kj 4859.280 kj 1 ml (, leic acid) -842.28 kj/ml = 800 kj/ml =, leic acid ALE 28 - Page 5 5