In the reaction, H2(g)+I2(g)⇌2HI(g), in a 2 litre flask 0.4 mole of each H2 and I2 are taken. At equilibrium 0.5 mol of HI are formed. What will be the value of equilibrium constant Kc?
A
20.2
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B
25.4
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C
0.284
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D
11.1
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Solution
The correct option is D11.1 In the reaction, H2(g)+I2(g)⇌2HI(g), in a 2 litre flask 0.4 mole of each H2 and I2 are taken. At equilibrium 0.5 mol of HI are formed. The value of equilibrium constant Kc will be 11.1. 0.5 moles of HI will be obtained from 0.25 moles of H2 and 0.25 moles of I2. 0.4−0.25=0.15 moles of H2 and 0.4−0.25=0.15 moles of I2 will remain. The equilibrium constant
Kc=[HI]2[H2][I2]
Kc=( 0.5 mol 2 L )2( 0.15 mol 2 L )×( 0.15 mol 2 L )