The variation of equilibrium constant K of a certain reaction with temperature T is ln K -3-0-2-0 - 104- T given R -8-3 J K -1 mol -1- the values of - H 0 and - S 0 are-A- - H -166 kJ - molB- - H 0-166 kJ - molC- - S 0-24-9 kJ - molD- - S 0-24-9 kJ - mol

Name: JEE - Grade 11 - Chemistry - GOC -1 - Session 12 - W14
Uploaded: 2022-07-04T10:36:42+05:30
Description: The correct options are B Δ H^o=+166 kJ/mol C Δ S^o=+24.9 kJ/molGiven, ln K=3.0+2.0×10^4T On differentiating w.r.t. T δ ln Kδ T=2.0×10^4T^2... (1) Also, δ ln ...

The correct options are B Δ H^o=+166 kJ/mol C Δ S^o=+24.9 kJ/molGiven, ln K=3.0+2.0×10^4T On differentiating w.r.t. T δ ln Kδ T=2.0×10^4T^2... (1) Also, δ ln ...

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Standard XII

Chemistry

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

The variation...

Question

The variation of equilibrium constant ( $K$ ) of a certain reaction with temperature ( $T$ ) is $ln K = 3.0 + \frac{2.0 \times 10^{4}}{T}$ given $R = 8.3 J K^{- 1} m o l^{- 1}$ , the values of $Δ H^{o}$ and $Δ S^{o}$ are:

Δ H^{o} = - 166 k J / m o l

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Δ H^{o} = + 166 k J / m o l

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Δ S^{o} = + 24.9 k J / m o l

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Δ S^{o} = - 24.9 k J / m o l

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Solution

The correct options are
B $Δ H^{o} = + 166 k J / m o l$
C $Δ S^{o} = + 24.9 k J / m o l$
Given,
$ln K = 3.0 + \frac{2.0 \times 10^{4}}{T}$
On differentiating w.r.t. $T$
$\frac{δ l n K}{δ T} = \frac{2.0 \times 10^{4}}{T^{2}} . . . (1)$
Also,
$\frac{δ l n K}{δ T} = \frac{Δ H^{o}}{R T^{2}} . . . (2)$
Equating $(1)$ and $(2)$
$Δ H^{o} = 2.0 \times 10^{4} \times R$
$Δ H^{o} = 2.0 \times 10^{4} \times 8.3$
$Δ H^{o} = + 166 k J / m o l$
Further,
$Δ G^{o} = - R T l n K$
$Δ G^{o} = - R T (3.0 + \frac{2.0 \times 10^{4}}{T})$
$Δ G^{o} = - 3 R T - 2.0 \times 10^{4} R$
Also,
$(\frac{δ Δ G^{o}}{δ T})_{p} = - Δ S^{o}$
$Δ S^{o} = 3 R$
$Δ S^{o} = 3 \times 8.3$
$Δ S^{o} = + 24.9 J m o l^{- 1} K^{- 1}$

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

Standard XII Chemistry

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The variation of equilibrium constant (K) of a certain reaction with temperature (T) is lnK=3.0+2.0×104T given R=8.3 JK−1mol−1, the values of ΔHo and ΔSo are:

The variation of equilibrium constant ( $K$ ) of a certain reaction with temperature ( $T$ ) is $ln K = 3.0 + \frac{2.0 \times 10^{4}}{T}$ given $R = 8.3 J K^{- 1} m o l^{- 1}$ , the values of $Δ H^{o}$ and $Δ S^{o}$ are: