The value of log10K for a reaction A - B is- Given- r H298 K- - -54-07 kJ mol-1- r S298 K- - 10 J K-1 mol-1 and R -8-314 J K-1 mol-1

Given: _r H_298 K^∘ = 54.07 kJ mol^ 1, _r S_298 K^∘ = 10 J K^ 1 mol^ 1, R =8.314 J K^ 1 mol^ 1, T=298 K We know the relation G^∘= H^∘ T S^∘ At equilib ...

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Standard XIII

Chemistry

Enthalpy of Combustion

The value of ...

Question

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

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Solution

The correct option is A 10
Given: $△_{r} H_{298 K}^{\circ} = - 54.07 k J m o l^{- 1}, △_{r} S_{298 K}^{\circ} = 10 J K^{- 1} m o l^{- 1}, R = 8.314 J K^{- 1} m o l^{- 1}, T = 298 K$
We know the relation $△ G^{\circ} = △ H^{\circ} - T △ S^{\circ}$
At equilibrium $△ G^{\circ} = 0$
$0 = - 54.07 \times 1000 - 298 \times 10 0 = - 57050 J m o l^{- 1} \Rightarrow - 57050 = - 5705 {log}_{10} K {log}_{10} K = 10$

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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

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 (Given:

△_{r} H_{298 k} = - 54.07 k J . m o l^{- 1},

△_{r} S_{298 k}^{\circ} = + 10 J . m o l^{- 1}

, and

R = 8.31 J . K^{- 1} m o l^{- 1})

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:
(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}

)

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Enthalpy of Combustion

Standard XIII Chemistry

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The value of log10K for a reaction A⇌B is: Given:△rH∘298 K=−54.07 kJ mol−1,△rS∘298 K=10 J K−1 mol−1 and R=8.314 J K−1 mol−1

The correct option is A 10Given: △rH∘298 K=−54.07 kJ mol−1,△rS∘298 K=10 J K−1 mol−1, R=8.314 J K−1 mol−1, T=298 K We know the relation △G∘=△H∘−T△S∘ At equilibrium △G∘=0 0=−54.07×1000−298×100=−57050 J mol−1⇒−57050=−5705log10Klog10K=10

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