NH4Cl(s) → NH3(g) + HCl(g) ΔH°=176 kJ/mol
At 25 °C, this reaction has a ΔG° of 91.1 kJ/mol.
What is the ΔS° of this reaction?
I came up with the solution of 0.896 kJ/mol. Is this correct?
halo!!!!! the answer i got correct was 0.285 kJ/K hope dis helps the 764 ppl who were lookin for the answer to dis question <3
-0.896 kJ/Mol isn't right... Sorry don't know the answer :/
I don't get that. If you will post your work I will check it. Did you remember that dG and dH are in kJ/mol and S is in J/mol?
yeah, it's definitely 0.285 kJ/K
Well, you did a great job there! But let's break it down a bit more.
To find the ΔS° of the reaction, we can use the equation ΔG° = ΔH° - TΔS°, where T is the temperature in Kelvin. Since the given temperature is 25 °C, we need to convert it to Kelvin first.
25 °C + 273.15 = 298.15 K
Now, we'll rearrange the equation to solve for ΔS°:
ΔS° = (ΔH° - ΔG°) / T
ΔS° = (176 kJ/mol - 91.1 kJ/mol) / 298.15 K
ΔS° = 84.9 kJ/mol / 298.15 K
Calculating that, we find ΔS° to be approximately 0.285 kJ/mol. So it seems there was a little calculation mistake there, but hey, we all make mistakes! Keep up the good work!
To find the ΔS° (change in entropy) of the reaction, you can use the equation:
ΔG° = ΔH° - TΔS°
Where:
ΔG° is the standard Gibbs free energy change
ΔH° is the standard enthalpy change
T is the temperature in Kelvin
ΔS° is the standard entropy change
Given:
ΔG° = 91.1 kJ/mol
ΔH° = 176 kJ/mol
T = 25°C = 25 + 273.15 = 298.15 K (temperature in Kelvin)
Now, let's rearrange the equation to solve for ΔS°:
ΔS° = (ΔH° - ΔG°) / T
Substituting the given values:
ΔS° = (176 kJ/mol - 91.1 kJ/mol) / 298.15 K
ΔS° = 84.9 kJ/mol / 298.15 K
ΔS° ≈ 0.285 kJ/(mol·K)
Therefore, the correct value for ΔS° of this reaction is approximately 0.285 kJ/(mol·K). So, it seems that your calculated value of 0.896 kJ/mol is not correct.