Question

For each of the following scenarios, use a calculation to predict whether or not it will occur spontaneously.

5.0 L of an ideal gas (C_v, m = 5/2 R) in a balloon at 1.5 atm, 298 K undergoes isothermal expansion against 1.0 atm until the final pressure of the gas equals the external pressure. The combustion reaction for glucose at 298 K carried out in a bomb calorimeter. Sublimation of 0.50 mol of dry ice at 277 K (T_sub = 277 K, delta H degree _sub = 25 kJ/mol) that is left out in a room held at 298 K.

Answer 1

moles of gas containing 5 L volume at 1.5 atm pressure and 298 K

n= PV/RT = 5*1.5/(0.0821* 298)=0.31

Entropy chaange for isothermal expansion = -nR ln (P2/P1)

where P2= 1 atm and P1= 1.5 atm. entropy change = -0.31* 8.314*ln(1/1.5) = 4.15J/K

for isothernal expansion deltaH=0

deltG= deltaH- TdeltaS= - 298*4.15 Joules=-1237 Joules since this is -ve, the process is spontaneous

2. since enthalpy change is -ve for combustion of glcuose, deltaH is -ve.

for glucose combustion there is an increase in entropy

hence deltaG= deltaH- TdeltaS and hence combustion of glucose is spontaneous

3. entropy change of dry ice = 0.5* 25 Kj/mole/277 = 0.045 KJ/K

Entropy change of room = -0.5*25/298 =-0.042 KJ/K

total entropy change = 0.045+0.042= 0.003 Kj/K

deltaG for sublimation is +ve

hence deltaG = deltaH- TdeltaS

0.5*25-298*0.003 =11.605 Kj

the process is not spontaneous