The value of log10 K for a reaction A- B is- Given - - rH0298k-54-07 kJ mol-1- - rS0298k-10JK-1mol-1 and R-8-314 JK-1mol-1- 2-303- 8-314- 298-5705

The correct option is B 10 Δ G^0=Δ H^0 TΔ S^0= 54.07 × 1000 298 × 10 = 54070 2980= 57050 Δ G^0= 2.303 RT log_10K 57050= 2.303× 298× 8.314 ...

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Gibbs Free Energy & Spontaneity

The value of ...

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The value of $l o g_{10}$ K for a reaction $A ⇌ B$ is:
$(G i v e n : Δ_{r} H_{298 k}^{0} = - 54.07 k J m o l^{- 1},$ $Δ_{r} S_{298 k}^{0} = 10 J K^{- 1} m o l^{- 1}$ $a n d R = 8.314 J K^{- 1} m o l^{- 1};$
$2.303 \times 8.314 \times 298 = 5705)$

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l o g_{10} K

A ⇌ B

Δ_{r} H_{298 K}^{0} = - 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}

2.303 \times 8.314 \times 298 = 5705 J 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

△_{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) .

{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}

log K

A (g) ⇌ B (g)

Δ_{r} H_{298 k}^{o} = - 54.07 k J m o l^{- 1}, Δ_{r} S_{298 K}^{o} = - 191.62 k J / K

2.303 \times 8.314 \times 298 = 5705

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