Thermal decomposition of gaseous X2 to gaseous X at 298 K takes place according to the following equation- X2g - 2XgThe standard reaction Gibbs energy- - Go- of this reaction is positive- At the start of the reaction- there is one mole of X2 and no X- As the reaction proceeds- the number of moles of X formed is given by - Thus- -equilibrium 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 mol-1The equilibrium constant Kp for this reaction at 298 K- in terms of -equilibrium- is

X_2 → 2X(g) G^o gt; 0 1 0 1 β_eq/2 β_eq RT InK_eq gt; 0 ...

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Byju's Answer

Standard XII

Chemistry

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 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

$\frac{8 β_{e q u i l i b r i u m}^{2}}{2 - β_{e q u i l i b r i u m}}$

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$\frac{8 β_{e q u i l i b r i u m}^{2}}{4 - β_{e q u i l i b r i u m}^{2}}$

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$\frac{4 β_{e q u i l i b r i u m}^{2}}{2 - β_{e q u i l i b r i u m}}$

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$\frac{4 β_{e q u i l i b r i u m}^{2}}{4 - β_{e q u i l i b r i u m}^{2}}$

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Solution

The correct option is B
$\frac{8 β_{e q u i l i b r i u m}^{2}}{4 - β_{e q u i l i b r i u m}^{2}}$

$X_{2} \to 2 X (g)$ $△ G^{o} > 0$
1 0
$1 - β_{e q / 2}$ $β_{e q}$ $- R T I n K_{e q} > 0$
$R T I n K_{e q} < 0$
$2.303 R T l o g (\frac{β^{2}}{1 - β^{2} / 2^{2}}) . 2 < 0$
$k_{p} = \frac{2 β^{2}}{1 - β^{2} / 4} = \frac{8 β^{2}}{4 - β^{e q u i l i b r i u m}}$

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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 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

Q. 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:

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. The equilibrium constant

K_{p}

for the homogeneous gaseous reaction is

10^{- 3}

. The standard Gibbs free energy change

Δ G^{o}

for the reaction at

27^{o} C

(sing

R = 2 c a l K^{- 1} {m o l}^{- 1}

) is_____________.

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The correct option is B 8β2equilibrium4−β2equilibrium X2 → 2X(g) △Go>0 1 0 1−βeq/2 βeq −RTInKeq>0 RTInKeq<0 2.303RT log (β21−β2/22).2<0 kp=2β21−β2/4=8β24−βequilibrium