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Standard XII

Chemistry

Concentration Terms - An Introduction (w/w etc)

In a 5.76I ...

Question

In a $5.76 I$ vessel, $0.5$ moles of $H_{2}$ gas and $0.5$ moles of $I_{2}$ vapours are allowed to react to form $H I (g)$ at $447^{o} C$ , then total pressure of gases at equilibrium would be
$(R = 0.08 l - a t m / K - m o l)$

20 a t m

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10 a t m

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5 a t m

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1 a t m

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Solution

The correct option is B $20 a t m$
$H_{2} + I_{2} ⇌ 2 H I$
$V = 5.76 L; R = 0.08 L a t m / K m o l$
$P V = n R T$
$\Rightarrow P (5.76) = (\frac{1}{2}) (0.08) (720)$
$\Rightarrow P = 20 a t m$

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In a 5.76I vessel, 0.5 moles of H2 gas and 0.5 moles of I2 vapours are allowed to react to form HI(g) at 447oC, then total pressure of gases at equilibrium would be(R=0.08l−atm/K−mol)

The correct option is B 20atmH2+I2⇌2HIV=5.76L;R=0.08 L atm/K molPV=nRT⇒P(5.76)=(12)(0.08)(720)⇒P=20 atm

In a $5.76 I$ vessel, $0.5$ moles of $H_{2}$ gas and $0.5$ moles of $I_{2}$ vapours are allowed to react to form $H I (g)$ at $447^{o} C$ , then total pressure of gases at equilibrium would be
$(R = 0.08 l - a t m / K - m o l)$

The correct option is B $20 a t m$
$H_{2} + I_{2} ⇌ 2 H I$
$V = 5.76 L; R = 0.08 L a t m / K m o l$
$P V = n R T$
$\Rightarrow P (5.76) = (\frac{1}{2}) (0.08) (720)$
$\Rightarrow P = 20 a t m$