Chapter 2, Exercises #2.25(b), 2.26(b), 2.29(b). 2.34(b), 2.35(b), 2.36(b), 2.39(b), 2.42(b), 2.43(b); Problem 2.2. Data from Tables 2.5 and 2.6 (in text Appendix).
 

2.25(b)
    

2.26(b)
    

2.29(b)
    
 

2.34(b)

AgBr(s) Ag⁺(aq) + Br^-(aq)

2.35(b)

C(gr) + O₂(g) CO₂(g) H₁ = -393.51 kJ/mol
 

C(dia) + O₂(g) CO₂(g) H₂ = -395.41 kJ/mol
 

By subtraction: C(gr) C(dia) H₃ = H₁ - H₂ = -393.51-(-395.41) = 1.9kJ
 

2.36(b)

for glucose: M = 180.16g/mol; H_comb = -2808 kJ/mol
 

Therefore, for 2.5g, (2.5g/180.16g/mol)(-2808kJ/mol) = 38.97kJ
 

w = mgh = (38.97x10³J)(0.25) = (65kg)(9.81m/s²)h
 

h = 15.28m
 

2.39(b)a) C₃H₆(cyclopropane) C₃H₆(propene)
 

H_r = H_f(propene) - H_f(cyclopropane) = 20.42 - 53.3 = -32.9kJ
 

b) HCl(aq) + NaOH(aq) NaCl(aq) + H₂O(l)
 

For strong electrolytes, the reaction is just H⁺ (aq) + OH^-(aq) H₂O(l)
 

H_r = H_f(H₂O)- H_f(OH^-) - H_f(H⁺) = -285.83 - (-229.99) - (0) = -55.84kJ
 

2.42(b) 2NOCl(g) 2NO(g) + Cl₂(g)
 

H_r = 2H_f(NO) + H_f(Cl₂) - 2H_f(NOCl)
 

H_f(NOCl) = 0.5[2(90.25) + 0 - 75.5] = 52.5kJ
 

2.43(b)
    
 

2.2 For water _p = 75.291J/mol-K and H_vap = 44.016kJ/mol at 298K
 

The equivalent mass of water = 65kg 65x10³g/18g/mol = 3.611x10³mol
 

For the input of 10MJ per day of heat energy,
 

T = q/C_p = 10x10⁶J/(3.611x10³mol)(75.291J/mol-K) = 36.78K = 36.78C = 66.21F
 

Need to absorb the 10MJ by evaporation, H₂O(liq) H₂O(g)
 

which requires n = (10MJ)/(44.016kJ/mol) = 227.2 moles
 

and m = (227.2mol)(18g/mol) = 4.09kg