Calculate Δ f G ∘ for (NH 4 Cl, s) at 310 K. Given : Δ f H ∘ NH 4 Cl, s = - 314.5 kJ/mol ; Δ r C p = 0 S N 2 g ∘ = 192 JK -1 mol -1 ; S H 2 g ∘ = 130.5 JK -1 mol -1 ; S Cl 2 g ∘ = 233 JK -1 mol -1 ; S NH 4 Cl s ∘ = 99.5 JK -1 mol -1 All given data at 300 K.

Step-by-step solution

∆ f S 0 N H 4 C l , s a t 300 K = S N H 4 C l ( S ) 0 − [ 1 2 S N 2 0 + 2 S H 2 0 + 1 2 S C l 2 0 ] = - 374 J K - 1 m o l - 1 ∴ ∆ f C p = 0 ∴ ∆ f S 310 0 = ∆ r S 300 0 = - 374 J K - 1 m o l - 1 ∆ f H 310 0 = ∆ f H 300 0 = - 314.5 ∆ f G 310 0 = ∆ f H 0 - 310 ∆ f S 0 = - 314.5 - 310 - 374 1000 - 198.56 k J / m o l