The value of log10 K for a reaction A - B isGiven- -r H298k0 - -54-07 kJ-1mol-1- 2-303 - 8-314 - 298 - 5705

A ⇌ B Δ G^0 = Δ H^0 TΔ S^0 Δ G^0 = 2.303 RT log_10 K 2.303 RT log_10 K = TΔ S^0 Δ H^0 log_10K = TΔ S^0 Δ H^0/2.303 RT = 298 × 10 + 54.07 × 1000/2.3 ...

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Standard XIII

Chemistry

Gibbs Free Energy & Spontaneity

The value of ...

Question

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

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

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Solution

The correct option is B
10

$A ⇌ B$

$Δ G^{0} = Δ H^{0} - T Δ S^{0}$

$Δ G^{0} = - 2.303 R T l o g_{10} K$

$- 2.303 R T l o g_{10} K = T Δ S^{0} - Δ H^{0}$

$l o g_{10} K = \frac{T Δ S^{0} - Δ H^{0}}{2.303 R T} = \frac{298 \times 10 + 54.07 \times 1000}{2.303 \times 8.314 \times 298}$

= $10$

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Similar questions

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

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

Q. 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}^{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) .

Q. 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)

Q. 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}^{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) .

Q. The value of

{log}_{10} K

for a reaction

A ⟶ B

is :

(G i v e n : Δ_{r} H_{298 K}^{o} = - 54.07 k J m o 1^{- 1}, Δ_{r} S_{298 K}^{o} = 10 J K^{- 1} m o 1^{- 1} a n d

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

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The value of log10 K for a reaction A ⇌ B is (Given: ΔrH0298k = −54.07 kJ−1mol−1;2.303 × 8.314 × 298 = 5705)

The correct option is B 10 A ⇌ B ΔG0 = ΔH0 − TΔS0 ΔG0 = −2.303 RT log10 K −2.303 RT log10 K = TΔS0 − ΔH0 log10K = TΔS0 − ΔH02.303 RT = 298 × 10 + 54.07 × 10002.303 × 8.314 × 298 = 10

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

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