The value of log10K for a reaction A- B is-Given- r H298K - -54-07 kJ mol-1- rSo298K-10 J K-1 mol-1 and R - 8-314 J K-1mol-1- 2-303 - 8-314 - 298 - 5705-

Given _r H_298K = 54.07 kJ mol^ 1 nbsp; _rS^o_298K=10 J K^ 1 mol^ 1 R = 8.314 J K^ 1mol^ 1 2.303 × 8.314 × 298 = 5705) we know G^o = H^o T S^o = 54. ...

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Standard XII

Chemistry

Gibbs Free Energy & Spontaneity

The value of ...

Question

The value of $l o g_{10}$ K for a reaction $A ⇌ B$ is:
Given: $△_{r} H_{298 K} = - 54.07 k J m o l^{- 1}, △_{r} S_{298 K}^{o} = 10 J K^{- 1} m o l^{- 1} a n d R = 8.314 J K^{- 1} m o l^{- 1}; 2.303 \times 8.314 \times 298 = 5705) .$

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Solution

The correct option is B 10
Given-
$△_{r} H_{298 K} = - 54.07 k J m o l^{- 1}$
$△_{r} S_{298 K}^{o} = 10 J K^{- 1} m o l^{- 1}$
$R = 8.314 J K^{- 1} m o l^{- 1}$
$2.303 \times 8.314 \times 298 = 5705)$
we know-
$△ G^{o} = △ H^{o} - T △ S^{o}$
$= - 54.07 \times 10^{3} - (298) (10) J m o l^{- 1}$
$= - 57050 J m o l^{- 1}$
We know,
$△ G^{o} = - 2.303 R T l o g K^{o}$
$l o g_{10} K^{o} = \frac{△ G^{o}}{2.303 R T} = \frac{57050 J m o l^{- 1}}{5705 J m o l^{- 1}} = 10$
hence, $l o g_{10} K^{o} = 10$

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Similar questions

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:

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}

and

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

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

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}

; and

2.303 \times 8.314 \times 298 = 5705 J m o l^{- 1}

)

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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Gibbs Free Energy & Spontaneity

Standard XII Chemistry

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The value of log10K for a reaction A⇌B is: Given: △r H298K=−54.07 kJ mol−1,△rSo298K=10 J K−1 mol−1 and R=8.314 J K−1mol−1; 2.303×8.314×298=5705).

The value of $l o g_{10}$ K for a reaction $A ⇌ B$ is:
Given: $△_{r} H_{298 K} = - 54.07 k J m o l^{- 1}, △_{r} S_{298 K}^{o} = 10 J K^{- 1} m o l^{- 1} a n d R = 8.314 J K^{- 1} m o l^{- 1}; 2.303 \times 8.314 \times 298 = 5705) .$