Question

46g of I_2 and 1g of H_2 are heated to attain equilibrium : H_2 + I_2 ⇌2HI ; Equilibrium mixture contains 1.9g of I_2. Calculate K_c.

Hint :- Answer must be 143.74​

Answer 1

$\Large{\textsf{\textbf{\underline{\underline{Given\::}}}}}$

Initially,

• 46 g of I₂ & 1 g of H₂ are heated to attain equilibrium.

At equilibrium,

• Mixture contains 1.9 g of I₂.

$\Large{\textsf{\textbf{\underline{\underline{To\:do\::}}}}}$

• The value $\rm{K_c}$.

$\Large{\textsf{\textbf{\underline{\underline{Solution\::}}}}}$

Given balanced chemical reaction is,

$\red\star\:\bf{H_2\:(g)\:+\:I_2\:(g)\:\rightarrow\:2\:H I\:(g)\:}$

☛ First of all we calculate the initial moles of H₂ & I₂.

For H₂ :

➠ Mass of H₂ = 2 × 1 = 2 g

And,

➠ 1 mole of H₂ is reacted here.

That means,

$\longrightarrow\:\tt{2\:g\:of\:H_2\: contains\:1\:mole\:}$

According to the question,

• 1 g of H₂ is reacted in this reaction.

$\longrightarrow\:\tt{1\:g\:of\:H_2\: contains\:\left(\dfrac{1}{2}\right)\:moles\:}$

$\longrightarrow\:\bf{1\:g\:of\:H_2\: contains\:(0.5)\:moles\:}$

Hence,

✯ Initial mole of H₂ is 0.5 mole.

For I₂ :

➠ Mass of I₂ = 2 × 126 = 252 g

And,

➠ 1 mole of I₂ is reacted here.

That means,

$\longrightarrow\:\tt{252\:g\:of\:I_2\: contains\:1\:mole\:}$

$\longrightarrow\:\tt{1\:g\:of\:I_2\: contains\:\left(\dfrac{1}{252}\right)\:moles\:}$

According to the question,

• 46 g of I₂ is reacted in this reaction.

$\longrightarrow\:\tt{46\:g\:of\:I_2\: contains\:\left(\dfrac{1}{252}\times{46}\right)\:moles\:}$

$\longrightarrow\:\bf{46\:g\:of\:I_2\: contains\:(0.1825)\:moles\:}$

Hence,

✯ Initial mole of I₂ is 0.1825 mole.

⠀

✔ Now we arrange the reaction as,

⠀⠀⠀⠀⠀$\red\star\:\bf{H_2\:(g)\:\:+\:\:\:I_2\:(g)\:\:\rightarrow\:\:2\:H I\:(g)}$

Initially : 0.5 moles ⠀0.1825moles⠀⠀0

$\bf{At\atop{equilibrium}}\::$ (0.5 - x)⠀⠀(0.1825 - x) ⠀ 2x

[NOTE ➝ If at equilibrium x moles of ⠀⠀⠀⠀⠀⠀each of H₂ & I₂ have reacted, ⠀⠀⠀⠀⠀⠀then 2x moles of HI will be ⠀⠀⠀⠀⠀⠀formed.]

⠀

According to the question,

• At equilibrium I₂ contains 1.9 g.

$\longrightarrow\:\tt{1.9\:g\:of\:I_2\: contains\:\left(\dfrac{1}{252}\times{1.9}\right)\:mole\:}$

$\longrightarrow\:\bf{1.9\:g\:of\:I_2\: contains\:(0.0075)\:moles\:}$

From the above reaction,

• At equilibrium I₂ contains (0.1825 - x) moles.

$:\implies\:\tt{0.1825\: -\: x\:=\:0.0075\:}$

$:\implies\:\tt{x\:=\:0.1825\:-\:0.0075\:}$

$:\implies\:\bf{x\:=\:0.175\:moles}$

Hence,

✯ At equilibrium 0.175 moles of each H₂ & I₂ have reacted, so 0.35 moles of HI will be formed.

As we know that,

❂ According to the law of mass action, for synthesis of HI, the formula for $\rm{K_c}$ is

$\orange\bigstar\:{\Large\mid}\:\bf\blue{K_c\:=\:\dfrac{4x^2}{(a\:-\:x)\:(b\:-\:x)}\:}\:{\Large\mid}\:\green\bigstar$

Where,

• a & b are the initial moles of H₂ & I₂ respectively.

• x is the moles that have reacted from each H₂ & I₂ at equilibrium.

Here,

• a = 0.5 moles ⠀⠀

• b = 0.1825 moles⠀

• x = 0.175 moles

$\dashrightarrow\:\tt{K_c\:=\:\dfrac{4\times{(0.175)^2}}{(0.5\:-\:0.175)\:(0.1825\:-\:0.175)}\:}$

$\dashrightarrow\:\tt{K_c\:=\:\dfrac{0.1225}{0.325\times{0.0075}}\:}$

$\dashrightarrow\:\tt{K_c\:=\:\dfrac{0.1225}{0.0024}\:}$

$\dashrightarrow\:{\bf{\pink{K_c\:=\:51.041\:mol.L^{-1}\:}}}$

∴ The value of $\bf{K_c}$ is $\bf\gray{51.041\:mol.L^{-1}\:}$.

Answer 2

Given :-

• 46g of l₂ and 1 g of H₂ are heated to attain equilibrium.
• Mixture contains 1.9 g of l₂

Solution :-

Given balanced chemical reaction is,

$\sf \dag \: H₂(g)+I₂(g)→2HI(g)$

➺ 1st of all we calculate the initial moles of H₂ and 1₂

For H₂

➺ Mass of H₂ = 2 × 1 = 2g

➺ 1 mole of H₂ is reacted here .

⇢2g of H₂ contains 1 mole.

According to the question :

• 1g of H₂ is reacted in this reaction.

$\sf⇢1g of \: H₂ \: contains \: ( \frac{1}{2} )moles$

$\sf \: 1g \: of \: H₂ \: contains \: (0.5)moles$

Then :

✦ Initial mole of H₂ is 0.5 mole

For l₂ :

➟ Mass of l₂ = 2 × 126 = 252g

➟ 1 mole of l₂ is reacted here.

Its Means :

➺ 252 g of l₂ contains 1 moles

➺ 1g of l₂ contains (1/252) moles

According to questions,

• 46g of l₂ is reacted in this reaction.

⇒ 46g of l₂ contains (1/252 × 46)

⇒ 46g of l₂ contains (0.1825)m

Then :

★ Initial mole of l₂ is 0.1825 moles

✔ Now we arrange the reaction as,

✼ H₂ (g) + l₂ (g) ⇀2 HI (g)

Initially : 0.5 moles 0.1825 moles.