Given ∆rU∘=−10.5kJ ∆rS∘=+44.1J∕K ∆rG∘=? Using formula ∆H∘=∆U∘+∆(pV) ∆H∘=∆U∘+p∆V+V∆p as ∆p=0 So, ∆H∘=∆U∘+p∆V Now, as ∆G∘=∆H∘−T∆S∘ ∆G∘=∆U∘+p∆V−T∆S∘ p∆V=−RT ∵p∆V=∆ngRT ∆ng=(2−3)=−1 So, ∆G∘=∆U∘−RT−T∆S∘ =−10.5−(8.314×10−3×298)−(298×44.1×10−3) ≈−26.119kJ