Determine the entropy change when 1.80 mol of HBr(g) condenses at atmospheric pressure?

I got that DeltaS fus = 12.922J/(k*mol)
Delta S Vapour = 93.475J/(K*mol)
the only thing I don't get is the entropy change when 1.80 mol of HBr(g) condenses at atmospheric pressure. How do I do it?! I got an answer of -23.2596J/K because I multiplied the DeltaS fuion by the molar ratio and flipped the sign (-12.922*1.80) and got the wrong answer :-( The prompt says incorrect and "The value of ΔSvap above is for 1.00 mol of HBr. You need to scale that value up for 1.80 mol of HBr. Also, consider the sign of your answer. Does condensing result in an increase or decrease in entropy?" So I give up. anyone please help me get the value of DeltaS for HBr!
Extra Info;
Melting Point = -86.96 cel,
Delta h fusion = 2.406
boiling point = -67.0 cel,
delta h vapour = 19.27 kj/mol
(this is all the info they gave for HBr in the question) thank you in advance!

The entropy change when 1.80 mol of HBr(g) condenses at atmospheric pressure is -167.42 J/K. This is calculated by multiplying the molar entropy of vaporization (93.475 J/K*mol) by the number of moles (1.80 mol) and then subtracting the molar entropy of fusion (12.922 J/K*mol) multiplied by the number of moles (1.80 mol).

To determine the entropy change when 1.80 mol of HBr(g) condenses at atmospheric pressure, you need to consider the entropy changes during the phase transitions of HBr.

First, let's calculate the entropy change during the fusion (melting) of 1.80 mol of HBr. The molar ratio for fusion should be determined by the stoichiometry of the reaction. Since the prompt does not provide any specific reaction equation, we'll assume HBr(g) → HBr(l) for simplicity.

The equation for entropy change during fusion is:

ΔS_fus = ΔH_fus / T

where ΔH_fus is the enthalpy of fusion and T is the temperature in Kelvin.

Given the following information:
ΔH_fus = 2.406 kJ/mol
Melting Point of HBr = -86.96 °C = 186.19 K

Converting ΔH_fus to J/mol:
ΔH_fus = 2.406 kJ/mol x 1000 J/kJ = 2406 J/mol

Calculating ΔS_fus:
ΔS_fus = 2406 J/mol / 186.19 K = 12.948 J/(K*mol)

Next, let's determine the entropy change during the vaporization (boiling) of 1.80 mol of HBr. Similarly, assuming HBr(l) → HBr(g) for simplicity:

The equation for entropy change during vaporization is:

ΔS_vap = ΔH_vap / T

where ΔH_vap is the enthalpy of vaporization and T is the temperature in Kelvin.

Given the following information:
ΔH_vap = 19.27 kJ/mol
Boiling Point of HBr = -67.0 °C = 206.15 K

Converting ΔH_vap to J/mol:
ΔH_vap = 19.27 kJ/mol x 1000 J/kJ = 19270 J/mol

Calculating ΔS_vap:
ΔS_vap = 19270 J/mol / 206.15 K = 93.475 J/(K*mol)

Now we need to scale up the value of ΔS_vap for 1.80 mol of HBr. Since ΔS is an extensive property, meaning it scales with the amount of substance, we can simply multiply ΔS_vap by the number of moles (1.80 mol):

ΔS_vap_1.80mol = 93.475 J/(K*mol) x 1.80 mol = 168.255 J/K

Finally, to determine the entropy change during the condensation of 1.80 mol of HBr, we need to consider that condensation is the reverse process of vaporization. Therefore, the sign of the entropy change will be opposite to that of ΔS_vap.

Thus, the entropy change during the condensation of 1.80 mol of HBr at atmospheric pressure is:

ΔS_condensation = -ΔS_vap_1.80mol = -168.255 J/K

Note that the negative sign indicates a decrease in entropy, as the gas condenses into a liquid.

To determine the entropy change when 1.8 mol of HBr(g) condenses at atmospheric pressure, you can use the equations:

ΔS_fus = ΔH_fus / T_fus
ΔS_vap = ΔH_vap / T_vap

Where:
ΔS_fus is the entropy change during fusion
ΔH_fus is the enthalpy change during fusion
T_fus is the temperature at which fusion occurs
ΔS_vap is the entropy change during vaporization
ΔH_vap is the enthalpy change during vaporization
T_vap is the temperature at which vaporization occurs

First, let's find the entropy change during fusion (condensation of vaporized HBr to liquid HBr):

ΔS_fus = ΔH_fus / T_fus
= 2.406 kJ/mol / (-86.96 °C + 273.15) K
= 2.406 kJ/mol / 186.19 K
= 12.93 J/(K*mol)

Next, let's find the entropy change during vaporization (vaporization of liquid HBr to gaseous HBr):

ΔS_vap = ΔH_vap / T_vap
= 19.27 kJ/mol / (-67.0 °C + 273.15) K
= 19.27 kJ/mol / 206.15 K
= 93.48 J/(K*mol)

Now, since the given value of ΔS_vap is for 1.00 mol of HBr, and we have 1.80 mol of HBr, we need to scale up the entropy change during vaporization:

ΔS_vap (1.80 mol HBr) = ΔS_vap (1.00 mol HBr) * (1.80 mol HBr / 1.00 mol HBr)
= 93.48 J/(K*mol) * (1.80 mol HBr / 1.00 mol HBr)
= 168.27 J/(K*mol)

Now, to find the entropy change when 1.80 mol of HBr(g) condenses, we need to consider whether condensing results in an increase or decrease in entropy. Since we are going from a gaseous state to a liquid state, we can expect a decrease in entropy. Therefore, the final entropy change is:

ΔS_final = - (ΔS_fus + ΔS_vap)
= - (12.93 J/(K*mol) + 168.27 J/(K*mol))
= - 181.20 J/(K*mol)

Hence, the entropy change when 1.80 mol of HBr(g) condenses at atmospheric pressure is -181.20 J/(K*mol).