Plz help me in this problem,i wud appreciate.

Question

Cn some1 plz help me in this problem.i wud appreciate.

the ΔG for the freezing of H2O(l) is -210J/mol & the heat of fusion of ice at this temperature is 5610J/mol.find the nethropy change of the universe when 1 mol of water freezes at -10°c.

Answer 1

Thank you for correcting me, my first answer couldn't of been any more wrong. The units for entropy are always in Kelvin.

Here is the solution for to 2 moles of water:

delta Gsystem = delta Hsystem - T(delta Ssystem)

(delta Hsystem - delta Gsystem)/T = delta Ssystem

First I converted the ethalpy of fusion into the enthalpy of freezing.Heres why:

This is kinda simple.

Fusion is:

H2O(s) + heat --> H2O(l)

Freezing is simply the reverse reaction:

H2O(l) --> heat + H2O(s)

Remember this rule of thumb the enthalpy change for the reverse reaction is the negative of the forward reaction, so:

heat of freezing = -(heat of fusion) = -(5610J/mol)

delta Hsys = (2mole) *(-5610J/mol) = -11220J

delta Gsys = (2mole) *(-210J/mol) = -420J

(-11220J - (-420J))/(263.15) = -10800J/K = delta Ssystem

delta Suniverse = delta Ssystem - (delta Hsystem)/T

delta Suniverse = -10800J/C - (-11220J)/263.15K

= -10757J/K or 10.757kJ/K

A negative sign means that this is an nonspontaneous process. Which kinda doesn't make sense because water freezes spontaneouly at 0C so it would be frozen at -10C. I think the reason why this may be nonspontaneous is that the pressure exerted on the system must be higher that 1atm. As the pressure exerted on the gas increases the lower the freezing point.

Thanks again for correcting me.