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?
To find the entropy change (ΔS°) of the reaction, we can use the Gibbs-Helmholtz equation, which relates ΔG° to ΔH° and ΔS°:
ΔG° = ΔH° - TΔS°
where:
ΔG° is the standard Gibbs free energy change of the reaction,
ΔH° is the standard enthalpy change of the reaction,
ΔS° is the standard entropy change of the reaction, and
T is the temperature in Kelvin.
In this case, we are given:
ΔG° = 91.1 kJ/mol
ΔH° = 176 kJ/mol
T = 25 °C = 298 K (temperature needs to be in Kelvin)
Plugging in these values into the equation, we get:
91.1 kJ/mol = 176 kJ/mol - 298 K * ΔS°
Rearranging the equation to solve for ΔS°, we get:
ΔS° = (176 kJ/mol - 91.1 kJ/mol) / 298 K
Calculating this value, we find:
ΔS° ≈ 0.289 J/(mol·K)
So, the ΔS° of this reaction is approximately 0.289 J/(mol·K).
To find the ΔS° (standard entropy change) 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, ΔS° is the standard entropy change, and T is the temperature in Kelvin.
Given that ΔG° = 91.1 kJ/mol and ΔH° = 176 kJ/mol, we can rearrange the equation and solve for ΔS°:
ΔS° = (ΔH° - ΔG°) / T
Since the temperature T is given as 25 °C, we need to convert it to Kelvin by adding 273.15:
T = 25 °C + 273.15 = 298.15 K
Now we can substitute the values into the equation:
ΔS° = (176 kJ/mol - 91.1 kJ/mol) / 298.15 K
Calculating the value:
ΔS° = 84.9 kJ/mol / 298.15 K
ΔS° ≈ 0.285 kJ/(mol·K)
Therefore, the ΔS° of the reaction is approximately 0.285 kJ/(mol·K).