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Equilibrium Constant and Standard Free Energy Change

Thermal decom...

Question

Thermal decomposition of gaseous $X_{2}$ to gaseous X at 298 K takes place according to the following equation:
$X_{2} (g) ⇌ 2 X (g)$
The standard reaction Gibbs energy, $Δ_{r} G^{0},$ of this reaction is positive. At the start of the reaction, there is one mole of $X_{2}$ and no X. As the reaction proceeds, the number of moles of X formed is given by $β$ . Thus, $β_{e q u i l i b r i u m}$ is the number of moles of X formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally. (Given: R = 0.083 L bar $K^{- 1} m o l^{- 1}$ ).

The INCORRECT statement among the following, for this reaction is:

Decrease in the total pressure will result in formation of more moles of gaseous X

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At the start of the reaction, dissociation of gaseous

X_{2}

takes place spontaneously

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β_{e q u i l i b r i u m} = 0.7

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K_{c} < 1

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Solution

The correct option is C $β_{e q u i l i b r i u m} = 0.7$
There is no data given to find the $β_{e q u i l i b r i u m}$ exact value.
$Δ G_{c}^{0} = - 2.303 R T l o g K_{c}$
$l o g K_{c} = - 1$
$K_{c} < 1$

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Similar questions

Q. Thermal decomposition of gaseous

X_{2}

to gaseous X at 298 K takes place according to the following equation:

X_{2} (g) ⇌ 2 X (g)

The standard reaction Gibbs energy,

Δ_{r} G^{0},

of this reaction is positive. At the start of the reaction, there is one mole of

X_{2}

and no X. As the reaction proceeds, the number of moles of X formed is given by

β

. Thus,

β_{e q u i l i b r i u m}

is the number of moles of X formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally. (Given: R = 0.083 L bar

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

).

The INCORRECT statement among the following, for this reaction is:

Q. Thermal decomposition of gaseous

X_{2}

to gaseous X at 298 K takes place according to the following equation:

X_{2} (g) ⇌ 2 X (g)

The standard reaction Gibbs energy,

Δ_{r} G^{0},

of this reaction is positive. At the start of the reaction, there is one mole of

X_{2}

and no X. As the reaction proceeds, the number of moles of X formed is given by

β

. Thus,

β_{e q u i l i b r i u m}

is the number of moles of X formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally. (Given: R = 0.083 L bar

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

)

The equilibrium constant

K_{p}

for this reaction at 298 K, in terms of

β_{e q u i l i b r i u m},

is:

Consider the following statements:

(I) An increase in pressure (caused by decrease in volume) at equilibrium results in increase in molar concentration of each gaseous substance involved.

(II) An increase in pressure (caused by decrease in volume) at equilibrium results in increase in no. of moles of each gaseous substance involved.

(III) For the reaction

A (s) ⇌ B (g) + C (g)

at constant temperature, total pressure at equilibrium (of

B (g)

C (g)

) is not affected by changing the volume of container.

(IV)

H_{2} O (l) ⇌ H_{2} O (g)

on temperature increase,

K_{p}

of this reaction is increased.

Tick the correct statement(s)

Q. The standard Gibbs energy change for a gaseous reaction at 27 is X Kcal. If equilibrium constant a reaction is 100 and R is 2 cal

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

. Then X is:

Q. For the gaseous reaction :

C_{2} H_{2} + D_{2} O ⇌ C_{2} D_{2} + H_{2} O

Δ H

530 c a l

. At

25 K_{p} = 0.82

. If X moles of

C_{2} D_{2}

will be formed, if 1 mole

C_{2} H_{2}

and 2 mole of

D_{2} O

are put together at a total pressure of 1 atm at

100

, then

100 X

is:

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The correct option is C βequilibrium=0.7There is no data given to find the βequilibrium exact value. ΔG0c=−2.303RT log Kc log Kc=−1 Kc<1

The correct option is C $β_{e q u i l i b r i u m} = 0.7$
There is no data given to find the $β_{e q u i l i b r i u m}$ exact value.
$Δ G_{c}^{0} = - 2.303 R T l o g K_{c}$
$l o g K_{c} = - 1$
$K_{c} < 1$