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Standard XII

Chemistry

Le Chatelier's Principle for Del N Equals to 0

At 46∘C, Kp...

Question

At $46^{\circ} C, K_{p}$ for the reaction $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ is 0.667 atm. Compute the percent dissociation of $N_{2} O_{4}$ at $46^{\circ} C$ at a total pressure fo 380 Torr.

50%

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70%

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40%

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20%

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Solution

The correct option is A 50%
$N_{2} O_{4} \to 2 N O_{2}$
At eqm 1-x $2 \times x$
$K_{p} = (2 x / 1 + x)^{2} \times (380 / 760)^{2} / (1 - x / 1 + x) \times (380 / 760) = 0.667.$
$x = 0.5.$
So 50% dissociation happens.

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Similar questions

Q. At

46

K_{p}

for the reaction

{N_{2} O}_{4} (g) ⇌ 2 N O_{2} (g)

0.667

atm. Compute the percent dissociation of

{N_{2} O}_{4}

46

at total pressure of

380

Torr.

Q. In a container equilibrium,

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

is attained at

25^{\circ} C

. The total equilibrium pressure in container is

380 torr

. If equilibrium constant of above equilibrium is

0.667 atm

, then degree of dissociation of

N_{2} O_{4}

at this temperature will be:

Q. In a container equilibrium

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

is attained at

25^{o} C .

The total equilibrium pressure in container is 380 torr. If equilibrium constant of above equilibrium is 0.667 atm, then degree of dissociation of

N_{2} O_{4}

at this temperature will be:

Q. For the dissociation reaction,

N_{2} O_{4} ⇌ 2 N O_{2} (g)

, the equilibrium constant

K_{p}

is 0.120 atm at 298 K and total pressure of system is 2 atm. Calculate the degree of dissociation of

N_{2} O_{4}

Q. In a container, equilibrium

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

is attained at

25^{\circ}

C. The total equilibrium pressure in the container is

380

torr. If equilibrium constant of the above equilibrium is

0.667 a t m

, then the degree of dissociation of

N_{2} O_{4}

at this temperature will be:

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At 46∘C,Kp for the reaction N2O4(g)⇌2NO2(g) is 0.667 atm. Compute the percent dissociation of N2O4 at 46∘C at a total pressure fo 380 Torr.

The correct option is A 50% N2O4→2NO2At eqm 1-x 2×xKp=(2x/1+x)2×(380/760)2/(1−x/1+x)×(380/760)=0.667.x=0.5.So 50% dissociation happens.

At $46^{\circ} C, K_{p}$ for the reaction $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ is 0.667 atm. Compute the percent dissociation of $N_{2} O_{4}$ at $46^{\circ} C$ at a total pressure fo 380 Torr.

The correct option is A 50%
$N_{2} O_{4} \to 2 N O_{2}$
At eqm 1-x $2 \times x$
$K_{p} = (2 x / 1 + x)^{2} \times (380 / 760)^{2} / (1 - x / 1 + x) \times (380 / 760) = 0.667.$
$x = 0.5.$
So 50% dissociation happens.