Question

To answer Questions 2.1-2.4, you will need the thermodynamic data below. SPECIES A | B | C | D. | AH1° 0 kJ/mol 0 kJ/mol | +185 kJ/mol -360 kJ/mol | Note that the letters should not be interpreted as elemental symbols (for example, "C" does not stand for "carbon"), but rather treated as variables standing in for species of unknown identity. 2.1. Calculate the enthalpy change for the reaction below. Express your answer in kJ/mol C rounded to the nearest whole number and be sure to indicate whether you answer is positive or negative. C --> B+ A

Answer 1

- > 8+ A srº - DHCProduct) - Soup Reactant). VE 23 2H = CH (B) + DHA (A) - 41, (C) = 0 + 0 - 185 kJ molt = -185 kJmolte B + 20 D cho = LHOLD) - DHCB2 - 2014 (C) = (-360-0-2x185) kJ md = -730 kJ mol-D Potential energy In 2.2, more of heat is released so potential energy of products is lower in enugy relative to products of 201. There is more change in P.E. in 2.2. Change in Kinetic Enegy & More enugy is released in 2.2 relative to 2.1. so products have less enugy and exhibit less motion, Temperature heat is released in both reactions so both make hand feel hot but 22 makes more hot. favorability. It is explained on basis of DG only. So it can't be concluded

I 2.4 AG = AH -TAS (VB+H AG = -185kJ melal TAS AH = ove for both Ren. B+2CD ; DG = -730 kJ mot -TAS. C B +A avourable at lowt only, i.e. when TAS is small so it means to that is is the here, ire. no. of gaseous species are to decreasing CB+A so Bng = 1 at least GB A a att so is gaseous and B & A are solid so entropy decreases. B+2C D 2 lave urable at all temperature so as is the so that she is negative at all temp. so Dis gaseous and B, C are solid, so possible phases = A = solid ? B = solid explains entropy changes well C = solid Da gaseous According to above conditions, as in 211 can be small regative which dominates at high temperature and make DG tve.