The value of - log K for a reaction - Ag - Bg-Given- - r Ho 298k - -54-07kJ mol-1- - r So298K-191-62kJ-K and 2-303- 8-314- 298 - 5705-

The correct option is A 10 We know RTlnK= Δ H T Δ S ⇒ 2.303 RT log K= Δ H T Δ S ⇒ 5705 log K= 54.07+(298 × 191.62) #160; #160; ...

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The value of ...

Question

The value of - $log K$ for a reaction :

$A (g) ⇌ B (g)$

[Given: $Δ_{r} H_{298 k}^{o} = - 54.07 k J m o l^{- 1}, Δ_{r} S_{298 K}^{o} = - 191.62 k J / K$ and $2.303 \times 8.314 \times 298 = 5705$ ]

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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)

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) .

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

(Given: $Δ_{r} H_{298 k}^{0} = - 54.07 k J^{- 1} m o l^{- 1}; 2.303 \times 8.314 \times 298 = 5705)$

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) .

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