What is the ΔrH°, ΔrS°, ΔrG° at 298.15 K for the reaction below.
Cl2(g) + Br2(l) ⟶ 2 BrCl(g)
ΔfH ° at 298.15 K 0 0 14.68
/ kJ mol−1
S¯ ∘ at 298.15 K 223.1 152.2 240.1
/ J K−1 mol−1
To determine the standard enthalpy change (ΔrH°), standard entropy change (ΔrS°), and standard Gibbs free energy change (ΔrG°) at 298.15 K for the given reaction, we will use the following equations:
ΔrH° = ΣΔfH°(products) - ΣΔfH°(reactants)
ΔrS° = ΣS°(products) - ΣS°(reactants)
ΔrG° = ΔrH° - TΔrS°
First, we need to obtain the values for the standard enthalpy of formation (ΔfH°) and standard entropy (S°) for each species involved in the reaction.
From the given table, we have:
ΔfH° at 298.15 K:
ΔfH°(Cl2) = 0 kJ mol−1
ΔfH°(Br2) = 0 kJ mol−1
ΔfH°(BrCl) = 14.68 kJ mol−1
S° at 298.15 K:
S°(Cl2) = 223.1 J K−1 mol−1
S°(Br2) = 152.2 J K−1 mol−1
S°(BrCl) = 240.1 J K−1 mol−1
Now, we can calculate the values required:
ΔrH°:
ΔrH° = ΣΔfH°(products) - ΣΔfH°(reactants)
= 2ΔfH°(BrCl) - (ΔfH°(Cl2) + ΔfH°(Br2))
= 2(14.68 kJ mol−1) - (0 kJ mol−1 + 0 kJ mol−1)
= 29.36 kJ mol−1
ΔrS°:
ΔrS° = ΣS°(products) - ΣS°(reactants)
= 2S°(BrCl) - (S°(Cl2) + S°(Br2))
= 2(240.1 J K−1 mol−1) - (223.1 J K−1 mol−1 + 152.2 J K−1 mol−1)
= 480.2 J K−1 mol−1 - 375.3 J K−1 mol−1
= 104.9 J K−1 mol−1
Now, we can calculate ΔrG° using the equation: ΔrG° = ΔrH° - TΔrS°
Given:
T = 298.15 K
ΔrG° = ΔrH° - TΔrS°
= 29.36 kJ mol−1 - (298.15 K)(104.9 J K−1 mol−1)(1 kJ / 1000 J)
= 29.36 kJ mol−1 - (31.2621175 kJ mol−1)
= -1.9021175 kJ mol−1
Therefore, the values for ΔrH°, ΔrS°, and ΔrG° at 298.15 K are as follows:
ΔrH° = 29.36 kJ mol−1
ΔrS° = 104.9 J K−1 mol−1
ΔrG° = -1.9021175 kJ mol−1
Note: The negative value of ΔrG° indicates that the reaction is spontaneous in the forward direction at 298.15 K.