The value of log10 K for a reaction Aleftharpoons B is (Given: Δ rHo298 K=-54.07 kJmol- 1,Δ rSo298 K=10J K- 1mol- 1 and R=8.314 JK- 1mol- 1;2.303× 8.314× 298=5705 )

Question

The value of log10 K for a reaction Aleftharpoons B is (Given: Δ rHo298 K=-54.07 kJmol- 1,Δ rSo298 K=10J K- 1mol- 1 and R=8.314 JK- 1mol- 1;2.303× 8.314× 298=5705 ) (A) 90 (B) 100 (C) 5 (D) 10

Tardigrade Admin · Accepted Answer

Δ Go=Δ Ho-TΔ So=-54.07× 1000-298× 10=-57050 Jmol- 1 Δ G° =-2.303RTlog10K -57050=-5705log10K log10K=10

Q. The value of $l o g_{10}$ K for a reaction $A ⇌ B$ is

(Given: $Δ_{r} H_{298 K}^{o} = - 54.07$ $k J m o l^{- 1}, Δ_{r} S_{298 K}^{o} = 10 J$ $K^{- 1} m o l^{- 1}$

and $R = 8.314$ $J K^{- 1} m o l^{- 1}; 2.303 \times 8.314 \times 298 = 5705$ )

90

100

5

10

$Δ G^{o} = Δ H^{o} - T Δ S^{o} = - 54.07 \times 1000 - 298 \times 10 = - 57050$ $J m o l^{- 1}$

$Δ G^{\circ} = - 2.303 RTl o g_{10} K$

$- 57050 = - 5705 l o g_{10} K$

$l o g_{10} K = 10$

Q. The value of log10​ K for a reaction A⇌B is (Given: Δr​H298Ko​=−54.07 kJmol−1,Δr​S298Ko​=10J K−1mol−1 and R=8.314 JK−1mol−1;2.303×8.314×298=5705 )

90

100

5

10

ΔGo=ΔHo−TΔSo=−54.07×1000−298×10=−57050 Jmol−1 ΔG∘=−2.303RTlog10​K −57050=−5705log10​K log10​K=10

Q. The value of $l o g_{10}$ K for a reaction $A ⇌ B$ is

(Given: $Δ_{r} H_{298 K}^{o} = - 54.07$ $k J m o l^{- 1}, Δ_{r} S_{298 K}^{o} = 10 J$ $K^{- 1} m o l^{- 1}$

and $R = 8.314$ $J K^{- 1} m o l^{- 1}; 2.303 \times 8.314 \times 298 = 5705$ )

$Δ G^{o} = Δ H^{o} - T Δ S^{o} = - 54.07 \times 1000 - 298 \times 10 = - 57050$ $J m o l^{- 1}$

$Δ G^{\circ} = - 2.303 RTl o g_{10} K$

$- 57050 = - 5705 l o g_{10} K$

$l o g_{10} K = 10$