Examine the following reaction:
NH4Cl (s)→NH3 (g)+HCl (g)ΔH0=176 kJ
At 25 °C, this reaction has a ΔG0 of 91.1 kJ.
What is the ΔS0 of this reaction?
dGo = dHo - TdSo
91100 = 176,000 - 298^dSo
Solve for dSo in Joules.
It's, 0.285 kJ/K
To determine the ΔS0 (standard entropy change) of the given reaction, we can use the equation:
ΔG0 = ΔH0 - TΔS0
Where:
ΔG0 is the standard free energy change
ΔH0 is the standard enthalpy change
T is the temperature in Kelvin
ΔS0 is the standard entropy change
Given:
ΔG0 = 91.1 kJ
ΔH0 = 176 kJ
T = 25 °C = 298 K
We can rearrange the equation to solve for ΔS0:
ΔS0 = (ΔH0 - ΔG0) / T
Substituting the values into the equation:
ΔS0 = (176 kJ - 91.1 kJ) / 298 K
ΔS0 = 84.9 kJ / 298 K
Calculating the value:
ΔS0 ≈ 0.285 kJ/K
Therefore, the ΔS0 of the reaction is approximately 0.285 kJ/K.