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?
dGo = dHo - T(dSo)
The answer I ended up with is 0.05. Is that correct?
To find the ΔS° (change in entropy) of the reaction, we can use the relationship between ΔG° (standard Gibbs free energy change), ΔH° (standard enthalpy change), and ΔS°.
The relationship is defined by the equation:
ΔG° = ΔH° - TΔS°
Where:
- T is the temperature in Kelvin (°C + 273.15).
In this case, we know ΔG° is 91.1 kJ/mol, and ΔH° is 176 kJ/mol. We need to solve for ΔS°.
Let's rearrange the equation and solve for ΔS°:
ΔG° = ΔH° - TΔS°
Rearranging:
ΔS° = (ΔH° - ΔG°) / T
Plugging in the values:
ΔS° = (176 kJ/mol - 91.1 kJ/mol) / (25°C + 273.15)
Calculating:
ΔS° = 84.9 kJ/mol / 298.15 K
ΔS° ≈ 0.284 kJ/(mol·K)
Therefore, the ΔS° of the reaction is approximately 0.284 kJ/(mol·K).