Can someone help me with this? I was using deltarG = deltarH - T deltarS. I found delta H= 62438 j/mol and delta S =. 144.555 j/ mol k
Calculate ΔrG for the sublimation of iodine gas at 298 K when the iodine vapor is at a pressure of 0.0500 bar
62548 -298*1333.555=-33458joules/mole
Wrong!
To calculate ΔrG for the sublimation of iodine gas at 298 K when the iodine vapor is at a pressure of 0.0500 bar, you can follow these steps:
1. Convert the pressure from bar to pascals (Pa). Since 1 bar is equal to 100,000 Pa, multiply the given pressure by 100,000:
0.0500 bar * 100,000 Pa/bar = 5,000 Pa
2. Recall that ΔrG (change in Gibbs Free Energy) is given by the equation:
ΔrG = ΔrH - TΔrS
where ΔrH is the change in enthalpy, T is the temperature in Kelvin, and ΔrS is the change in entropy.
3. Substitute the known values into the equation:
ΔrH = 62438 J/mol
T = 298 K
ΔrS = 144.555 J/mol K
ΔrG = 62438 J/mol - (298 K * 144.555 J/mol K)
4. Calculate TΔrS:
TΔrS = 298 K * 144.555 J/mol K = 43,043.59 J/mol
5. Subtract TΔrS from ΔrH:
ΔrG = 62438 J/mol - 43,043.59 J/mol
6. Calculate the value of ΔrG:
ΔrG = 18,394.41 J/mol
Therefore, the value of ΔrG for the sublimation of iodine gas at 298 K when the iodine vapor is at a pressure of 0.0500 bar is 18,394.41 J/mol.