Pastpaper 95------chemistry

(a) (i) First of all, 2 moles of hydrogen gas are needed for formation of 1 mole of water.

Under 298K and 1 atm, the change in volume of hydrogen gas is:

△V = 2 × 8.31 × 298/(1.013 × 105) = 0.0489 m3

Assuming that the volume of water (liquid) is negligible compared to the hydrogen gas originally present, the work done is given by:

W = P△V
= 1.013 × 105 × 0.0489
= 4.95 kJ

(ii) Since the container volume is fixed, △V = 0 and hence the mechanical work done owing to expansion/collapse is zero.

So to speak, 140.3 kJ of energy is released when 1.00 g of hydrogen gas is burnt completely which is equivalent to 0.5 moles, i.e. 0.25 moles of water is formed.

Therefore, the enthalpy change of formation of water at 298K is given by:

140.3/0.25 = 561.2 kJ/mol


2007-12-28 16:54:44 補充：
Sorry for the mistake:For (i):When 1 mol of water is formed, the no. of moles of gases disappeared should be 1.5 moles (1 mole of H2 and 0.5 moles of O2)Therefore the work done should be:1.5 x 8.31 x 298 = 3.7 kJ

2007-12-28 16:58:32 補充：
(ii)Molar mass of H2 = 2.016 gTherefore when 1 mole of water is formed, △U = -2.016 x 140.3Also, under normal conditions, P△V is also considered and then, by first law of thermodynamics:△H = △U 十 p△V= -2.016 x 140.3 - 3.7= 286.5 kJ/mol