The value of log10 K for a reaction A - B is Given- rH298k-54-07 kJ- mol-1- rS-298k-10 J-mol-1- and R-8-31 J- K-1 mol-1

The correct option is A 10 A⇌ B , #160; Δ G^∘=Δ H^∘ TΔ S^∘ , Δ G^∘= 2.303 RT log K, log K= Δ H^∘ TΔ S^∘ 2.303 RT, log K= 54.0 ...

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Question

The value of $l o g_{10}$ K for a reaction $A ⇌ B$ is (Given: $△_{r} H_{298 k} = - 54.07 k J . m o l^{- 1},$ $△_{r} S_{298 k}^{\circ} = + 10 J . m o l^{- 1}$ , and $R = 8.31 J . K^{- 1} m o l^{- 1})$

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Similar questions

{log}_{10}

A ⇌ B

Given - △_{r} H_{298 K}^{\circ} = - 54.07 k J m o l^{- 1}, △_{r} S_{298 K}^{\circ} = 10 J K^{- 1} m o l^{- 1}

R - 8.314 J K^{- 1} m o l^{- 1}

l o g_{10}

A ⇌ B

△_{r} H_{298 K} = - 54.07 k J m o l^{- 1}, △_{r} S_{298 K}^{o} = 10 J K^{- 1} m o l^{- 1} a n d R = 8.314 J K^{- 1} m o l^{- 1}; 2.303 \times 8.314 \times 298 = 5705) .

l o g_{10}

A ⇌ B

(G i v e n : Δ_{r} H_{298 k}^{0} = - 54.07 k J m o l^{- 1},

Δ_{r} S_{298 k}^{0} = 10 J K^{- 1} m o l^{- 1}

a n d R = 8.314 J K^{- 1} m o l^{- 1};

2.303 \times 8.314 \times 298 = 5705)

{log}_{10} K

A ⇌ B

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}, R = 8.314 J K^{- 1} {m o l}^{- 1}

l o g_{10}

A ⇌ B

△_{r} H_{298 K} = - 54.07 k J m o l^{- 1}, △_{r} S_{298 K}^{o} = 10 J K^{- 1} m o l^{- 1} a n d R = 8.314 J K^{- 1} m o l^{- 1}; 2.303 \times 8.314 \times 298 = 5705) .

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