The variation of equilibrium constant K of a certain reaction with temperature T is ln K-3-0-2-0-104T given R-8-3 JK-1mol-1- the values of - Ho and - So are-

Given, 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 Kδ T=Δ H^oRT^2... (2) Equating (1) and nbsp;(2) Δ H^o= 2.0× ...

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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 option is 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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Similar questions

Q. 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:

Q. For the reaction,

X (s) ⇌ Y (s) + Z (g)

, the plot of

l n \frac{p_{z}}{p^{0}} versus \frac{10^{4}}{T}

is given below (in solid line), when

p_{z}

is the pressure (in bar) of the gas

Z

at temperature

T

and

p^{0} = 1

bar.

(Given,

\frac{d (l n K)}{d (\frac{1}{T})} = - \frac{Δ H^{0}}{R}

, where the equilibrium constant,

K = \frac{p_{z}}{p^{0}}

and the gas constant.

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

The value of

Δ S^{0}

(in

J K^{- 1} m o l^{- 1}

) for the given reaction, at

1000 K

Q. The equilibrium constant for the reaction is

10

. Calculate the value of

Δ G^{o}

; (Given

R = 8.314 {J K}^{- 1} {mol}^{- 1}; T = 300 K

Q. Variation of equilibrium constant

K

with temperature

T

is given by van't Hoff equation:

log K = log A - \frac{Δ H^{o}}{2.303 R T}

A graph between

log K

and

T^{- 1}

was a straight line as shown in the figure and having

θ = {tan}^{- 1} (- 0.5)

and

O P = 10

. Calculate:
(a)

Δ H^{o}

(Standard heat of reaction) when

T = 298 K

(b)

A

(pre-exponential factor)
(c) Equilibrium constant

K

298 K

(d)

K

798 K

, if

Δ H^{o}

is independent of temperature

Q. The equilibrium constant for a reaction is 10. what will be value of

△ G^{⊖}

? R =8.314

J K^{- 1} m o l^{- 1}

, T=300 K

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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: