A mixture of 0-3 mole of H2 and 0-3 mole of I2 is allowed to react in a 10 litre evacuated flask at 500 -C- The reaction is H2 - 12HI the K is found to be 64- The amount of unreacted 12 at equilibrium is -

The correct option is B 0.06 molesolution: #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; ...

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Byju's Answer

Standard XII

Chemistry

Reaction Quotient

A mixture of ...

Question

A mixture of 0.3 mole of $H_{2}$ and 0.3 mole of $I_{2}$ is allowed to react in a 10 litre evacuated flask at $500^{\circ} C .$ The reaction is $H_{2} + 12 H I$ the K is found to be 64. The amount of unreacted 12 at equilibrium is ?

0.1 mole

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0.06 mole

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0.03 mole

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0.02 mole

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Solution

The correct option is B 0.06 mole
solution:
$H_{2}$ + $I_{2}$ $\to 2 H I$
Initial 0.3 0.3 --
At Equilibrium (0.3−x) (0.3−x) 2x
$K_{C} = \frac{[H I]^{2}}{[H 2] [I 2]}$

$\to 64 = \frac{[\frac{2 x}{10}]}{[\frac{0.3 - x}{10}] [\frac{0.3 - x}{10}]}$

$60 x^{2} - 38.4 x + 5.76 = 0$

$x = 0.24$
Hence,moles of $I_{2}$ unreacted =0.3−0.24=0.06 mole.
hence the correct option :B

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A mixture of 0.3 mole of H2 and 0.3 mole of I2 is allowed to react in a 10 litre evacuated flask at 500∘C. The reaction is H2+12HI the K is found to be 64. The amount of unreacted 12 at equilibrium is ?

The correct option is B 0.06 molesolution: H2 + I2 →2HIInitial 0.3 0.3 --At Equilibrium (0.3−x) (0.3−x) 2xKC=[HI]2[H2][I2]→64=[2x10][0.3−x10][0.3−x10]60x2−38.4x+5.76=0x=0.24Hence,moles of I2 unreacted =0.3−0.24=0.06 mole. hence the correct option :B

A mixture of 0.3 mole of $H_{2}$ and 0.3 mole of $I_{2}$ is allowed to react in a 10 litre evacuated flask at $500^{\circ} C .$ The reaction is $H_{2} + 12 H I$ the K is found to be 64. The amount of unreacted 12 at equilibrium is ?