The value of log 10 K for a reaction A leftharpoons B is (Given: Δr H298 K=-54.07 kJ mol -1 and Δr S=10 J K-1 mol -1 .R =8.314 J K -1 mol -1 ; 2.303 × 8.314 × 298=5705)

Question

The value of log 10 K for a reaction A leftharpoons B is (Given: Δr H298 K=-54.07 kJ mol -1 and Δr S=10 J K-1 mol -1 .R =8.314 J K -1 mol -1 ; 2.303 × 8.314 × 298=5705) (A) 5 (B) 10 (C) 95 (D) 100

Tardigrade Admin · Accepted Answer

For the reaction A leftharpoons B we have Δ G °=Δ H °- T Δ S °=-54.07-298 × 10 =-54070-2980 J mol -1 =-57050 J mol -1 Δ G °=- RT In K =-2.303 RT log K -5705=2.303 × 8.314 × 298 log 10 K, -57050=-5705 log 10 K log 10 K=10

Q. The value of log10​K for a reaction A⇌B is (Given: Δr​H298K​=−54.07kJmol−1 and Δr​S=10JK−1mol−1 R=8.314JK−1mol−1;2.303×8.314×298=5705)

5

10

95

100

For the reaction A⇌B we have ΔG∘=ΔH∘−TΔS∘=−54.07−298×10 =−54070−2980Jmol−1 =−57050Jmol−1 ΔG∘=−RTInK=−2.303RTlogK −5705=2.303×8.314×298log10​K, −57050=−5705log10​K log10​K=10

Q. The value of $lo g_{10} K$ for a reaction $A ⇌ B$ is
(Given: $Δ_{r} H_{298 K} = - 54.07 k J m o l^{- 1}$ and $Δ_{r} S = 10 J K^{- 1} m o l^{- 1}$
$R = 8.314 J K^{- 1} m o l^{- 1}; 2.303 \times 8.314 \times 298 = 5705)$