The value of log 10 K for a reaction A - B is-Given - r H 298 K 0-54-07 kJ mol -1- - r S 298 K 0-10 JK -1 mol -1- R -8-314 JK -1 mol -1 - and 2-303 - 8-314 - 298-5705 Jmol -1A- 5B- 10C- 95D- 100

The correct option is B 10We know, Δ G^o=Δ H^o TΔ S^o ⇒Δ G^o = ( 54.07×10^3 298×10) Jmol^ 1 ⇒Δ G^o= 57050 Jmol^ 1 Again, Δ G^o= RTlnK ⇒ logK=Δ G^o/2.303RT n ...

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Standard XII

Chemistry

Vant Hoff's Equation and Effect of Temperature on Equilibrium Constant

The value of ...

Question

The value of $l o g_{10} K$ for a reaction $A ⇌ B$ is:
(Given $Δ_{r} H_{298 K}^{0} = - 54.07 k J m o l^{- 1}$ , $Δ_{r} S_{298 K}^{o} = 10 J K^{- 1} m o l^{- 1}$ , $R = 8.314 J K^{- 1} m o l^{- 1}$ ; and $2.303 \times 8.314 \times 298 = 5705 J m o l^{- 1}$ )

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Solution

The correct option is B 10
We know, $Δ G^{o} = Δ H^{o} - T Δ S^{o}$
$\Rightarrow Δ G^{o} = (- 54.07 \times 10^{3} - 298 \times 10) J m o l^{- 1}$
$\Rightarrow Δ G^{o} = - 57050 J m o l^{- 1}$
Again, $Δ G^{o} = - R T l n K$
$\Rightarrow l o g K = \frac{Δ G^{o}}{2.303 R T}$
$\Rightarrow l o g K = \frac{57050}{5705} = 10$

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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:

(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

{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:

△_{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 log10K for a reaction A⇌B is: (Given ΔrH0298 K=−54.07 kJmol−1,ΔrSo298 K=10 JK−1mol−1, R=8.314 JK−1mol−1; and 2.303×8.314×298=5705 Jmol−1)

The correct option is B 10We know, ΔGo=ΔHo−TΔSo ⇒ΔGo=(−54.07×103−298×10) Jmol−1 ⇒ΔGo=−57050 Jmol−1 Again, ΔGo=−RTlnK ⇒logK=ΔGo2.303RT ⇒logK=570505705=10

The value of $l o g_{10} K$ for a reaction $A ⇌ B$ is:
(Given $Δ_{r} H_{298 K}^{0} = - 54.07 k J m o l^{- 1}$ , $Δ_{r} S_{298 K}^{o} = 10 J K^{- 1} m o l^{- 1}$ , $R = 8.314 J K^{- 1} m o l^{- 1}$ ; and $2.303 \times 8.314 \times 298 = 5705 J m o l^{- 1}$ )