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

Question

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 (A) 2+2 α (B) 2 (C) 1+ α (D) 2- α

Tardigrade Admin · Accepted Answer

2HI(g) leftharpoons H2(g)+I2(g) In initial 2 mol 0 mol 0 mol At equilibrium (2-2α) mol α mol α mol So, at equilibrium total moles =2-2α+α+α =2-2α+2α=2

Q. If $α$ is the fraction of HI dissociated at equilibrium in the reaction, $2 H I (g) ⇌ H_{2} (g) + I_{2} (g)$ starting with the 2 moles of HI, then the total number of moles of reactants and products at equilibrium are

$2 + 2 α$

2

$1 + α$

$2 - α$

$2 H I (g) ⇌ H_{2} (g) + I_{2} (g)$
In initial $2 m o l 0 m o l 0 m o l$
At equilibrium $(2 - 2 α) m o l α m o l α m o l$
So, at equilibrium total moles
$= 2 - 2 α + α + α$
$= 2 - 2 α + 2 α = 2$

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

2+2α

2

1+α

2−α

2HI(g)⇌H2​(g)+I2​(g) In initial 2mol0mol0mol At equilibrium (2−2α)molαmolαmol So, at equilibrium total moles =2−2α+α+α =2−2α+2α=2

Q. If $α$ is the fraction of HI dissociated at equilibrium in the reaction, $2 H I (g) ⇌ H_{2} (g) + I_{2} (g)$ starting with the 2 moles of HI, then the total number of moles of reactants and products at equilibrium are

$2 + 2 α$

$1 + α$

$2 - α$

$2 H I (g) ⇌ H_{2} (g) + I_{2} (g)$
In initial $2 m o l 0 m o l 0 m o l$
At equilibrium $(2 - 2 α) m o l α m o l α m o l$
So, at equilibrium total moles
$= 2 - 2 α + α + α$
$= 2 - 2 α + 2 α = 2$