[kazeAaqib]

Hi
I've been doing some thermodynamics questions, but some have gotten me a bit confused. Appreciated if you could please look at them and see if I'm doing them correctly.

1) The solution becomes colder when a salt is dissolved in water in a spontaneous process. What does that tell you about the entropy change of the system (ΔS) ?

I know Its a spontaneous reaction so the ΔS of the universe will increase

ΔS =q_rev/T
does this equation mean that the if the heat exchange (q) is positive (endothermic) the ΔS will be positive as well?

2) 2 moles of  steam at 100°C are condensed at 100°C and the resulting water is cooled to 25°C. What is the entropy change for this process?
(The molar heat capacity of liquid water is 75.3 J mol1 K1, and the molar enthalpy of vaporisation of water is 40.8 kJ mol1 at 100°C).

I wasn't sure if i had the the right answer for this question, for example am i right in using the molar heat capacity and not specific heat capacity. I'm also not sure of the units involved.
ΔvapS = ΔvapH/bp
           = 40.8/373
           = -0.109.38 kJ
ΔS       =C lnTf/Ti
           =73.5 x ln(25/100)
           = -101.89 
Total entropy for the condensation and heat loss equals both of the above
ΔS      =-0.109.38+-101.89
          = -211.27 J/K
This is multiplied by 2 to give 2 moles ( Im actually not to sure if the equation above gives molar ΔS)
            211.27 x 2 = 422.55

(3) When 2 moles of di-n-propyl ketone (C7H14O (l)) is completely burned in a bomb calorimeter at 298K, 8790 kJ of heat is evolved.
(iii) Calculate the molar enthalpy of combustion of di-n-propyl ketone.

I know the ΔU of the is -8790 kJ but am not too sure about the enthalpy of combustion

ΔH = ΔU + Δ (pv)
     = ΔU + (Δn) RT
     = 8790 + -3 x 8.314 x 298
     = -1357.28 kJ
This is then divided by 2 to give molar enthalpy of combustion
     = 678.64 kJ
 
Regarding the last question what if I had the calorimetry constant (70.5), T which is 298K and standard molar enthalpy of combustion and needed to find the standard change of internal energy ΔSΘ. Could just use the standard ΔH of combustion with the values of the bomb calorimeter to find this.