Question

1.When 35.0 mL of 1.43 M NaOH at 22.0^oC are neutralized by 35.0 mL of HCl also at 22.0^oC in a coffee cup calorimeter, the temperature of the final solution rises to 31.29^oC. The specific heat of all solutions and the density of all solutions is the same as that of water. Calculate the heat of neutralization for one mole of HCl.

2.

1.Given the following equation for the decomposition of aluminum oxide:                   

2 Al₂O₃(s) ---> 4 Al (s) + 3 O₂ (g) Delta (D)H^o = 3351.4 kJ

           What is Delta (D)H^o for the formation of 25.50 grams of aluminum oxide?

Answer 1

1.

Number of moles of HCl used = number of moles of NaOH = molarity × volume (L)

                                                 = 1.43 M × 35.0/1000 L = 0.05005 mol

Mass of neutralized mixture = (V1 + V2)/density = (35.0 + 35.0)/1 = 70.0 g

Heat generated (Q) = m . s . (T_f-T_i) = 70.0 g × (4.186 J/g°C) × (31.29 – 22.0) °C = 2713.3652 J

Heat of neutralization = Heat generated/number of moles = 2713.3652 J/0.05005 = 54213.091 J mol^-1

                                                                                                                                   = 54.21 kJ mol^-1

2.

2 Al2O3(s)   --->          4 Al (s) + 3 O2 (g)          Delta (D)H^o = 3351.4 kJ

The Delta (D)H^o for the reverse reaction = -3351.4 kJ

4 Al (s) + 3 O2 (g) --->      2 Al2O3(s)               Delta (D)H^o = -3351.4 kJ

Molar mass of Al2O3 = 101.96 g/mol

From the balanced equation,

Delta (D)H^o for the formation of 2 x 101.96 g Al2O3 = -3351.4 kJ

So, Delta (D)H^o for the formation of 25.50 g Al2O3 = -3351.4 x 25.50/(2 x 101.96) = 419.09 kJ