The value of log!0 K for a reaction A⇔B is (Given : Δ4H298K∘=−54.07kJmol−1,ΔrS298K∘=10JK−1mol−1 and R=8.314JK−1mol−1: 2.303×8.314×298=5705)

Name: Video solution 1: The value of log!0 K for a reaction A⇔B is (Given : Δ4H298K∘=−54.07kJmol−1,ΔrS298K∘=10JK−1mol−1 and R=8.314JK−1mol−1: 2.303×8.314×298=5705)
Uploaded: 2021-01-01T18:41:02.000Z
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Description: Video solution 1: The value of log!0 K for a reaction A⇔B is (Given : Δ4H298K∘=−54.07kJmol−1,ΔrS298K∘=10JK−1mol−1 and R=8.314JK−1mol−1: 2.303×8.314×298=5705)

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The value of {log}_{{!{0}}} K for a reaction {A}{B} is (Given

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The value of $log_{! 0}$ K for a reaction $A \Leftrightarrow B$ is (Given : $Δ_{4} H_{298 K}^{\circ} = - 54.07 k J mol^{- 1}, Δ_{r} S_{298 K}^{\circ} = 10 J K^{- 1} mol^{- 1}$ and R=8.314 $J K^{- 1} mol^{- 1}$ : 2.303 $\times 8.314 \times 298 = 5705$ )

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The value of

log_{! 0}

K for a reaction

A \Leftrightarrow B

is (Given :

Δ_{4} H_{298 K}^{\circ} = - 54.07 k J mol^{- 1}, Δ_{r} S_{298 K}^{\circ} = 10 J K^{- 1} mol^{- 1}

and R=8.314

J K^{- 1} mol^{- 1}

: 2.303

\times 8.314 \times 298 = 5705

)

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Question Text	The value of $log_{! 0}$ K for a reaction $A \Leftrightarrow B$ is (Given : $Δ_{4} H_{298 K}^{\circ} = - 54.07 k J mol^{- 1}, Δ_{r} S_{298 K}^{\circ} = 10 J K^{- 1} mol^{- 1}$ and R=8.314 $J K^{- 1} mol^{- 1}$ : 2.303 $\times 8.314 \times 298 = 5705$ )
Answer Type	Video solution: 1
Upvotes	153

The value of log!0​ K for a reaction A⇔B is (Given : Δ4​H298K∘​=−54.07kJmol−1,Δr​S298K∘​=10JK−1mol−1 and R=8.314JK−1mol−1: 2.303×8.314×298=5705)

Filo tutor solutions (1)

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The value of $log_{! 0}$ K for a reaction $A \Leftrightarrow B$ is (Given : $Δ_{4} H_{298 K}^{\circ} = - 54.07 k J mol^{- 1}, Δ_{r} S_{298 K}^{\circ} = 10 J K^{- 1} mol^{- 1}$ and R=8.314 $J K^{- 1} mol^{- 1}$ : 2.303 $\times 8.314 \times 298 = 5705$ )