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Standard XII

Chemistry

Equilibrium Constant and Standard Free Energy Change

Ammonia gas a...

Question

Ammonia gas at 15 atm is introduced in a rigid vessel at 300 K. At equilibrium the total pressure of the vessel is found to be 40.11 atm at 300 C. The degree of dissociation of $N H_{3}$ will be :

0.6

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0.4

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unpredictable

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none of these

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Solution

The correct option is B $0.4$
$2 N H_{3} ⇌ N_{2} + 3 H_{2}$
$T = 300 K$ 15 atm
$T = 573 K$ $\frac{15}{300} (573) a t m$
$\frac{15}{300} (573) [1 + α) = 40.11 \Rightarrow α = 0.4$

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Similar questions

Q. Ammonia gas at

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Ammonia gas at 15 atm is introduced in a rigid vessel at 300 K. At equilibrium the total pressure of the vessel is found to be 40.11 atm at 300 C. The degree of dissociation of NH3 will be :

The correct option is B 0.4 2NH3⇌N2+3H2T=300K 15 atmT=573K 15300(573)atm15300(573)[1+α)=40.11⇒α=0.4

Ammonia gas at 15 atm is introduced in a rigid vessel at 300 K. At equilibrium the total pressure of the vessel is found to be 40.11 atm at 300 C. The degree of dissociation of $N H_{3}$ will be :

The correct option is B $0.4$
$2 N H_{3} ⇌ N_{2} + 3 H_{2}$
$T = 300 K$ 15 atm
$T = 573 K$ $\frac{15}{300} (573) a t m$
$\frac{15}{300} (573) [1 + α) = 40.11 \Rightarrow α = 0.4$