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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, $Δ G^{o},$ of this reaction is positive. At the start of the reaction, there is one moe 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
$2.303 \times 0.083 \times 298 l o g (k_{p}) < 0$
$\frac{8 β^{2}}{4 - β^{2}} < 1$
$9 β^{2} < 4$
$β^{2} < 4 / 9$
$β < 2 / 3$
$k_{c} \times R T < 1$
$K_{c} < 1 / R T$
$K_{c} < 1$

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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. If the degree of dissociation of

P C l_{5}

in the reaction

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

at equilibrium is 0.2 and if the initial number of moles of

P C l_{5}

is one,the total number of moles present in the gaseous mixture at equilibrium is :

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The correct option is C βequilibrium = 0.72.303×0.083×298 log (kp)<0 8β24−β2<1 9β2<4 β2<4/9 β<2/3 kc×RT<1 Kc<1/RT Kc<1

The correct option is C $β_{e q u i l i b r i u m}$ = 0.7
$2.303 \times 0.083 \times 298 l o g (k_{p}) < 0$
$\frac{8 β^{2}}{4 - β^{2}} < 1$
$9 β^{2} < 4$
$β^{2} < 4 / 9$
$β < 2 / 3$
$k_{c} \times R T < 1$
$K_{c} < 1 / R T$
$K_{c} < 1$