Question

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

Answer 1

H_2+i_2 = 2HI

>H_2 = 1 g

>I_2 = 46 g

>2 × H × 1.9

>After finding H

>H = 87.4

>Kc = (product of molor concentration of product) ÷ (product of molar

Concentration of reactants)

=46÷166.06

=0.27

Answer 2

★ Concept :-

Here the concept of Equilibrium Constant has been used. We see that we are given the initial amount of the reactants and final mole of one of the product. So firstly we can find the initial moles of the substances. The we can find the final moles of the substances. The assuming the volume to be a constant, we can find molarity of the reactants and products and thus find the answer.

Let's do it !!

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★ Formula Used :-

$\;\boxed{\sf{k_{c}\;=\;\dfrac{[C][D]}{[A][B]}}}$

where A and B are reactants in the reaction and C and D are products in the reaction.

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★ Solution :-

Given,

» Intial mass of Iodine (I₂) = 46 g

» Initial mass of Hydrogen (H₂) = 1 g

» Mixture (product) at Equilibrium has mass of Iodine as = 1.9 g

» Molar Mass of I₂ = 253.8 g

» Molar Mass of H₂ = 2 g

$\;\bf{\red{\mapsto\;\;H_{2}\:+\:I_{2}\:\leftrightharpoons\;2HI}}$

• For the Intial Moles of the Reactants ::

$\;\tt{\leadsto\;\;No.\;of\;Moles\;=\;\dfrac{Given\;Mass}{Molar\;Mass}}$

>> Moles of H₂ ::

$\;\tt{\leadsto\;\;No.\;of\;Moles\;of\;H_{2}\;=\;\dfrac{1}{2}}$

$\;\;\bf{\leadsto\;\;No.\;of\;Moles\;of\;H_{2}\;=\;0.5}$

>> Moles of I₂ ::

$\;\tt{\leadsto\;\;No.\;of\;Moles\;of\;I_{2}\;=\;\dfrac{46}{253.8}}$

$\;\;\bf{\leadsto\;\;No.\;of\;Moles\;of\;I_{2}\;=\;0.1812}$

• For the Final Moles of Products ::

>> Moles of I₂ ::

$\;\tt{\leadsto\;\;No.\;of\;Moles\;of\;I_{2}\;=\;\dfrac{1.9}{253.8}}$

$\;\;\bf{\leadsto\;\;No.\;of\;Moles\;of\;I_{2}\;=\;0.0075}$

So number of moles converted into products from reactants is given as,

→ Final number of moles = 0.1812 - 0.0074 = 0.1737 moles

>> Moles of H₂ ::

Since the reactants and products are at Equilibrium. So the difference in final moles of Iodine and initial moles of Hydrogen will give final moles of Hydrogen.

→ Number of Moles of H₂ = 0.5 - 0.1737 = 0.32

The ratio of reactants and product is 1:2 . So the moles of product will be 2 × moles of Iodine.

>> Moles of HI ::

→ No. of moles of HI = 2 × 0.1737 = 0.35

• For the equilibrium constant ::

• Moles of I₂ = 0.0075

• Moles of H₂ = 0.32

• Moles of HI = 0.35

Now let's assume that the reaction takes place in the presence of a Volume of Solvent. So, this will be constant. Thus concentration of each substance will be divided by V.

We know the formula that,

$\;\bf{\rightarrow\;\;k_{c}\;=\;\dfrac{[C][D]}{[A][B]}}$

• Here terms in square bracket shows concentration of specific terms.

By applying values, we get

$\;\bf{\rightarrow\;\;k_{c}\;=\;\dfrac{[HI]^{2}}{[H_{2}][I_{2}]}}$

• Let the equilibrium constant of the reaction be Kc

Now putting the values of concentration we get,

$\;\bf{\rightarrow\;\;k_{c}\;=\;\dfrac{\bigg(\dfrac{0.35}{V}\bigg)^{2}}{\bigg(\dfrac{0.32}{V}\bigg)\bigg(\dfrac{0.0075}{V}\bigg)}}$

$\;\bf{\rightarrow\;\;k_{c}\;=\;\dfrac{\dfrac{(0.35)^{2}}{V^{2}}}{\dfrac{0.32\:\times\:0.0075}{V^{2}}}}$

$\;\bf{\rightarrow\;\;k_{c}\;=\;\dfrac{0.1225\:\times\:V^{2}}{0.0024\:\times\:V^{2}}}$

Cancelling V², we get

$\;\bf{\rightarrow\;\;k_{c}\;=\;\dfrac{0.1225}{0.0024}}$

$\;\bf{\rightarrow\;\;k_{c}\;=\;51.04\;\approx\;51}$

This is the required answer.

$\;\underline{\boxed{\tt{Equilibrium\;\: Constant \;\:of\;\:the\;\:reaction\;=\;\bf{\purple{51}}}}}$