Thermochemistry?responseToken=709e9d41724c2a3a23c702e98ec2f7af


 * November 12, 2007 (5.4)** Page 329

//**#4)**// //C2H2(g) + 2H2(g) -> C2H6(g)// //∆ H for 200g ethyne?// 1) C2H2(g) + 5/2O2 -> 2CO2(g) + H2O(l) ∆ H = -1299 kJ 2) H2 + ½ O2(g) --> H2O(l) ∆ H = -286 kJ 3) C2H6(g) + 7/2O2(g) -> 2CO2 (g) + 3H2O(l) ∆ H = -1560 kJ

a) 2 x (2) b) Reverse (3) c) (1) stays the same

a) 2H2 + O2(g) -> 2H2O(l) -572 kJ b) 3H2O(l) + 2CO2(g) -> 7/2O2 + C2H6(g) +1560 kJ c) C2H6(g) + 7/2O2(g) -> 2CO2 (g) + 3H2O(l) -1299 kJ C2H2(g) + 2H2(g) -> C2H6(g) + 311 kJ Molar mass of ethyne= 2(12) + 2 (1) = 26 g/mol 200g (1mol/26g) = 7.7mol ∆ H=(7.7mol)(311kJ/mol)=**2392.3 kJ**

//3H2 + CO (g)// //->////CH4 + H2O (g) ∆ H = ? for 300g CO// (1) 2H2(g) + O2(g) -> 2H2O(G) ∆ H = -483.6 kJ (2) 2C(s) + O2(g) -> 2CO(g) ∆ H = -221.0 kJ (3) CH4 +2O2(g) -> CO2(g) +2H2O(g) ∆ H = -802.7 kJ (4) C(s) + O2(g) -> CO2(g) ∆ H = -393.5 kJ (a) MULTIPLY (1) BY 3/2 (b) REVERSE (3) (c) REVERSE AND MULTIPLY (2) BY ½ (d) STAYS THE SAME Molar mass of carbon monoxide=12 + 1628g/mol 300g(1mol/28g)=10.7 mol ∆ H=(10.7 mol)(205.7 kJ/mol)=2200**.99 kJ
 * 5)**
 * 3H2 + CO (g)** **->** **CH4 + H2O (g) + 205.7 kJ** ∆ H = ? for 300g CO

Page 330**

1.a. (i) C8H18 + 25/2 O2(g) -> 8CO2(g) + 9H2O (g) + 5.47 MJ (ii) H2(g) + 1/2O2 -> H2O(l) + 285.8 kJ (iii) C + O2 -> CO2 + 393.5 kJ/mol b) 8C + 9H2 -> C8H18 (l) (1) 8 TIMES (iii) (2) 9 TIMES (ii) (3) REVERSE (i) ∆ **H = -250.2kJ**

2) HCl(g) + NaNO2(s) -> HNO2(g) + NaCl(s) ∆ H=? (i) 2NaCl + H2O -> 2HCl + Na2O ∆ H=507 kJ (ii) NO + NO2 +Na2O -> 2NaNO2 ∆ H = -427 kJ (iv) 2HNO2 -> N2O + O2 + H2O ∆ H = 34 kJ (a) Reverse (i) and multiply by ½ (b) Reverse (ii) and multiply by ½ (c) Reverse (iv) and multiply by ½ (d) Multiply (iii) by ½
 * ∆ H = -78.5kJ**

(i) Combustion of Ethanol: C2H5OH + 3O2 -> 2CO2 + 3H2O + 1367 kJ/mol (ii) Combustion of acetic acid: CH3COOH + 2O2 -> 2CO2 + 2H2O + 875 kJ/mol (a) C2H5OH + 3O2 -> 2CO2 + 3H2O ∆ H = - 1367 kJ/mol (b) 2CO2 + 2H2O -> CH3COOH + 2O2 ∆ H = 875 kJ/mol
 * 1) 3) C2H5OH + O2 -> CH3COOH+ H2O
 * C2H5OH + O2 -> CH3COOH+ H2O + 492 kJ/mol ∆ H = - 492 kJ/mol**

(1) HBr(aq) + KOH (aq) -> H2O(l) + KBr (aq) ∆ H = ? kJ (2) KOH (s) -> KOH (aq) ∆ H = ? kJ (3) KOH (s) + HBr (aq) -> H2O (l) + KBr(aq) ∆ H = ? kJ a) Add (1) and (2) to get (3): HBr(aq) + KOH (aq) + KOH (s) -> H2O(l) + KBr (aq)+ KOH (aq) b) c)
 * 4)