For the reaction, H2 + I2 leftharpoons 2HI, K= 47.6. If the initial number of moles of each reactant and product is 1 mole then at equilibrium

Question

For the reaction, H2 + I2 leftharpoons 2HI, K= 47.6. If the initial number of moles of each reactant and product is 1 mole then at equilibrium (A) [I2]=[H2], [I2] > [HI] (B) [I2]=[H2], [I2] < [HI] (C) [I2]<[H2], [I2] = [HI] (D) [I2]>[H2], [I2] = [HI]

Tardigrade Admin · Accepted Answer

For the given reaction, K = ([HI]2/[H2][I2]) As 1 mole of H2 reacts with 1 mole of I2, even at equilibrium, [H2] = [I2] Hence, K = ([HI]2/[I2]2) or √K=([HI]/[I2])=√47.6 i.e., [HI]>[I2]

Q. For the reaction, $H_{2} + I_{2} ⇌ 2 H I, K = 47.6.$ If the initial number of moles of each reactant and product is 1 mole then at equilibrium

$[I_{2}] = [H_{2}], [I_{2}] > [H I]$

$[I_{2}] = [H_{2}], [I_{2}] < [H I]$

$[I_{2}] < [H_{2}], [I_{2}] = [H I]$

$[I_{2}] > [H_{2}], [I_{2}] = [H I]$

For the given reaction, $K = \frac{[ H I ] ^{2}}{[ H _{2} ] [ I _{2} ]}$
As 1 mole of H $_{2}$ reacts with 1 mole of I $_{2}$ , even at equilibrium, $[H_{2}] = [I_{2}]$
Hence, $K = \frac{[ H I ] ^{2}}{[ I _{2} ] ^{2}}$ or $K = \frac{[ H I ]}{[ I _{2} ]} = 47.6$
i.e., $[H I] > [I_{2}]$

Q. For the reaction, H2​+I2​⇌2HI,K=47.6. If the initial number of moles of each reactant and product is 1 mole then at equilibrium

[I2​]=[H2​],[I2​]>[HI]

[I2​]=[H2​],[I2​]<[HI]

[I2​]<[H2​],[I2​]=[HI]

[I2​]>[H2​],[I2​]=[HI]