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A mixture of 0.3 mole of H(2) and 0.3 mo...

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^(@)`C. The reaction is `H_(2) + I_(2) hArr 2HI`, the K is found to be 64. The amount of unreacted `I_(2)` at equilibrium is

A

0.15 mole

B

0.06 mole

C

0.03 mole

D

0.2 mole

Text Solution

Verified by Experts

The correct Answer is:
B
`H_2(g) + I_(2)(g) hArr 2HI(g)`
`{:(0.3//10,0.3//10,0,"at start"),((0.3)/10-x/10,(0.3)/10-x/10,(2x)/10,"at equilibrium"):}`
`K_C = ((2x //10)^2)/(((0.3-x)/(10))^2)`
`64=(4x^2)/((0.3 -x)^2)`
`x=0.24`
Concentration of unreacted `I_2 = (0.3)/10 -x/10`
= 0.006 mole/L
Amount of unreacted `I_2 = 0.006 xx10 = 0.06 ` moles.
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