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Standard XII

Chemistry

Le Chatelier's Principle for Del N Equals to 0

Pure nitrosyl...

Question

Pure nitrosyl chloride $(N O C I)$ gas was heated to $240^{o} C$ in a $1.0 L$ container. At equilibrium the total pressure was $1.0 a t m$ and the $N O C l$ pressure was $0.64 a t m$ / What should be the value of $K_{p}$ ?

1.02 a t m

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16.875 \times 10^{- 3} a t m

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16 \times 10^{- 2} a t m

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N o n e o f t h e s e

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Solution

The correct option is B $16.875 \times 10^{- 3} a t m$
$2 N O C l (g) ⇌ 2 N O (g) + C l_{2} (g)$
At eqn $P - 2 x$ $2 x$ $x$
$P + x = 1$ ; $P - 2 x = 0.64$ , $x = 0.12$
$K_{p} = \frac{(0.24)^{2} (0.12)}{(0.64)^{2}}$
$= 16.875 \times 10^{- 3} a t m$

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At eqn	$P - 2 x$	$2 x$	$x$
$P + x = 1$ ;	$P - 2 x = 0.64$ ,	$x = 0.12$

Pure nitrosyl chloride (NOCI) gas was heated to 240oC in a 1.0 L container. At equilibrium the total pressure was 1.0 atm and the NOCl pressure was 0.64 atm/ What should be the value of Kp?

The correct option is B 16.875×10−3 atm2NOCl(g)⇌2NO(g)+Cl2(g)At eqn P−2x2xxP+x=1;P−2x=0.64,x=0.12Kp=(0.24)2(0.12)(0.64)2=16.875×10−3 atm

Pure nitrosyl chloride $(N O C I)$ gas was heated to $240^{o} C$ in a $1.0 L$ container. At equilibrium the total pressure was $1.0 a t m$ and the $N O C l$ pressure was $0.64 a t m$ / What should be the value of $K_{p}$ ?

The correct option is B $16.875 \times 10^{- 3} a t m$
$2 N O C l (g) ⇌ 2 N O (g) + C l_{2} (g)$
At eqn $P - 2 x$ $2 x$ $x$
$P + x = 1$ ; $P - 2 x = 0.64$ , $x = 0.12$
$K_{p} = \frac{(0.24)^{2} (0.12)}{(0.64)^{2}}$
$= 16.875 \times 10^{- 3} a t m$