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Standard XII

Chemistry

Characteristics of Chemical Equilibrium

At 1000 K, th...

Question

At 1000 K, the pressure of iodine gas is found to be 0.112 atm due to partial dissociation. If there has there been no dissociation the pressure would have been 0.074 atm. Calculate $K_{p}$ for the reaction, $I_{2} (g) \Leftrightarrow 2 I (g)$ .

1.60 atm

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16 atm

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0.016 atm

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0.160 atm

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Solution

The correct option is C 0.160 atm
$I_{2}$ $2 I$
Initial pressure $0.074$ $0$
Equilibrium pressure $0.074 - P$ $2 P$

The total pressure is $0.074 - P + 2 P = 0.112$ .
Hence, $P = 0.038$ .
The expression for the equilibrium constant will be $K_{p} = \frac{P_{I}^{2}}{P_{I_{2}}} = \frac{(2 \times 0.038)}{(0.074 - 0.038)} = 0.160 a t m$ .

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	$I_{2}$	$2 I$
Initial pressure	$0.074$	$0$
Equilibrium pressure	$0.074 - P$	$2 P$

At 1000 K, the pressure of iodine gas is found to be 0.112 atm due to partial dissociation. If there has there been no dissociation the pressure would have been 0.074 atm. Calculate Kp for the reaction, I2(g)⇔2I(g).

The correct option is C 0.160 atm I22IInitial pressure0.0740 Equilibrium pressure 0.074−P 2PThe total pressure is 0.074−P+2P=0.112. Hence, P=0.038.The expression for the equilibrium constant will be Kp=P2IPI2=(2×0.038)(0.074−0.038)=0.160 atm.

At 1000 K, the pressure of iodine gas is found to be 0.112 atm due to partial dissociation. If there has there been no dissociation the pressure would have been 0.074 atm. Calculate $K_{p}$ for the reaction, $I_{2} (g) \Leftrightarrow 2 I (g)$ .