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. ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

100

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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, $l o g_{10} K^{o} = 10$

Suggest Corrections

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

Join BYJU'S Learning Program

Explore more

Gibbs Free Energy & Spontaneity

Standard XII Chemistry

Join BYJU'S Learning Program

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