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  4. If α is the fraction of HI dissociated at equilibrium in the reaction, 2HI(g) leftharpoons H2(g)+I2(g) starting with the 2 moles of HI, then the total number of moles of reactants and products at equilibrium are

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Q. If $\alpha$ is the fraction of HI dissociated at equilibrium in the reaction, $ 2HI(g) \rightleftharpoons H_2(g)+I_2(g)$ starting with the 2 moles of HI, then the total number of moles of reactants and products at equilibrium are

IIT JEEIIT JEE 1996Aldehydes Ketones and Carboxylic Acids

Solution:

$2HI(g) \rightleftharpoons H_2(g)+I_2(g)$
In initial $ \, 2 \, mol 0 \, mol 0 \, mol$
At equilibrium $(2-2\alpha) mol \alpha \, mol \alpha \, mol$
So, at equilibrium total moles
$=2-2\alpha+\alpha+\alpha$
$=2-2\alpha+2\alpha=2$