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Byju's Answer

Standard XII

Chemistry

Derivation of Kp and Kc

if for the he...

Question

if for the heterogeneous equilibrium $C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g);$ K=1 at 1 atm, the temperature is given by:

T = \frac{Δ S^{0}}{Δ H^{0}}

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T = \frac{Δ H^{0}}{Δ S^{0}}

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T = \frac{Δ G^{0}}{R^{0}}

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T = \frac{Δ G^{0}}{Δ H^{0}}

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Solution

The correct option is B $T = \frac{Δ H^{0}}{Δ S^{0}}$
$Δ G = 2.303 R T l o g K$

As K =1 , $Δ G = 0$

We know the relation,

$Δ G = Δ H - T Δ S$

$T = \frac{Δ H}{Δ S}$

Option B is correct

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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. In conversion of limestone to lime,

C a C O_{3 (s)} \to C a O_{(s)} + C O_{2 (g)}

; the values of

Δ H^{0}

and

Δ S^{0}

are + 179.1 kJ/mol and 160.2J/K respectively, at 298K and 1 bar. Above what temperature, conversion of limestone to lime will be spontaneous (assuming that

Δ H^{0}

and

Δ S^{0}

do not change with temperature)?

In concersion of lime-stone to lime,

$C a C O_{3} (s) \to C a O (s) + C O_{2} (g)$ the values of $Δ H^{0}$ and $Δ S^{0}$ are $+ 179.1 k J m o l^{- 1}$ and $160.2 J / K$ respectively at 298 K and 1 bar. Assuming that $Δ H^{0}$ and $Δ S^{0}$ do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is

Q. Quick lime

(C a O)

is produced by heating limestone

(C a C O_{3})

to drive off

C O_{2}

gas.

C a C O_{3} (s) \to C a O (s) + C O_{2} (g), Δ H^{0} = 180 K J; Δ S^{0} = 150 J / K .

Assuming the variation of enthalpy change and entropy change with temperature to be negligible, choose the incorrect option(s).

Q. Calculate the equilibrium constant for the reaction given below at 400k.
If

Δ H^{0} = 77.2 k J m o l^{- 1}

and

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P C l_{5 (g)} \to P C l_{3 (g)} + C l_{2 (g)}

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if for the heterogeneous equilibrium CaCO3(s)⇌CaO(s)+CO2(g); K=1 at 1 atm, the temperature is given by:

The correct option is B T=ΔH0ΔS0ΔG=2.303RT logKAs K =1 , ΔG=0We know the relation,ΔG=ΔH−TΔST=ΔHΔSOption B is correct

if for the heterogeneous equilibrium $C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g);$ K=1 at 1 atm, the temperature is given by:

The correct option is B $T = \frac{Δ H^{0}}{Δ S^{0}}$
$Δ G = 2.303 R T l o g K$

As K =1 , $Δ G = 0$

We know the relation,

$Δ G = Δ H - T Δ S$

$T = \frac{Δ H}{Δ S}$

Option B is correct