Answer
Verified
394.5k+ views
Hint: The relation between reactant and product in reaction at the equilibrium state for a particular or specific unit can be expressed using equilibrium constant (k). It can give us information about the extent for a reaction i.e. the rate at which the reactants can disappear and the rate at which product can appear after the completion of a reaction.
Complete answer:
-Firstly, we have to write a balanced chemical equation:
${{\text{H}}_{2}}\text{ + }{{\text{I}}_{2}}\text{ }\to \text{ 2HI}$
-It is given that the volume of the flask is 1 litre, so the no. of moles is equal to the molar concentration.
-Also, it is given that the initial concentration of reactant and products are:
$\text{(}{{\text{H}}_{2}})\text{ = 1M, (}{{\text{I}}_{2}})\text{ = 2M and (HI) = 3M}$
-So, after the reaction completion, the reactant will disappear and the product will appear then, $({{\text{H}}_{2}})\text{ = 1 - x M, (}{{\text{I}}_{2}})\text{ }=\text{ 2 - x M and (HI) = 3 +2x M}$.
-Here. X is the concentration of hydrogen, iodine, hydrogen iodide.
-From the given reaction, the equilibrium constant will be:
${{\text{K}}_{\text{c}}}\text{ = }\dfrac{{{(\text{HI})}^{2}}}{({{\text{H}}_{2}})({{\text{I}}_{2}})}$
-It is given that the value of equilibrium constant id 45.9, so:
$\text{45}\text{.9 = }\dfrac{{{(3+2x)}^{2}}}{(1-x)(2-x)}$
$\text{45}\text{.9 = }\dfrac{9+4{{x}^{2}}+12x}{2-x-2x+{{x}^{2}}}\text{ = }\dfrac{9+4{{x}^{2}}+12x}{2-3x+{{x}^{2}}}$
$45.9(2-3x+{{x}^{2}})\text{ = 9+4}{{\text{x}}^{2}}+12x$ by solving it we will get,
$82.8-149.7x+41.9{{\text{x}}^{2}}\text{ = 0}$
Now, by using the quadratic equation we will get:
$x\text{ = }\dfrac{-\text{b }\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
$x\text{ = }\dfrac{-149.7\text{ }\pm \sqrt{{{(-149.7)}^{2}}-4(41.9)(82.8)}}{2(41.9)}$
$x\text{ = 2}\text{.888 or 0}\text{.684}$
-So, the value of x = 2.888 is not taken as the value of concentration will become negative. So, we will use x = 0.684.
-Hence, the equilibrium concentration will be:
$({{\text{H}}_{2}})\ \text{= 1-x = 1-0}\text{.684 = 0}\text{.316M}$
$({{\text{I}}_{2}})\ \text{= 2-x = 2-0}\text{.684 = 1}\text{.316M}$
$(\text{HI})\ \text{= 3 + 2x = 3 + 2(0}\text{.684) = 4}\text{.36M}$
Note: Le- chatelier Principle states that when a reaction is in equilibrium state then any change in the reaction due to change in the conditions will affect the rate of equation. In the answer, the concentration of hydrogen, iodine was subtracted from x because the concentration reactant decreases as the process starts whereas 2x was added with the concentration of HI because the concentration of product increases.
Complete answer:
-Firstly, we have to write a balanced chemical equation:
${{\text{H}}_{2}}\text{ + }{{\text{I}}_{2}}\text{ }\to \text{ 2HI}$
-It is given that the volume of the flask is 1 litre, so the no. of moles is equal to the molar concentration.
-Also, it is given that the initial concentration of reactant and products are:
$\text{(}{{\text{H}}_{2}})\text{ = 1M, (}{{\text{I}}_{2}})\text{ = 2M and (HI) = 3M}$
-So, after the reaction completion, the reactant will disappear and the product will appear then, $({{\text{H}}_{2}})\text{ = 1 - x M, (}{{\text{I}}_{2}})\text{ }=\text{ 2 - x M and (HI) = 3 +2x M}$.
-Here. X is the concentration of hydrogen, iodine, hydrogen iodide.
-From the given reaction, the equilibrium constant will be:
${{\text{K}}_{\text{c}}}\text{ = }\dfrac{{{(\text{HI})}^{2}}}{({{\text{H}}_{2}})({{\text{I}}_{2}})}$
-It is given that the value of equilibrium constant id 45.9, so:
$\text{45}\text{.9 = }\dfrac{{{(3+2x)}^{2}}}{(1-x)(2-x)}$
$\text{45}\text{.9 = }\dfrac{9+4{{x}^{2}}+12x}{2-x-2x+{{x}^{2}}}\text{ = }\dfrac{9+4{{x}^{2}}+12x}{2-3x+{{x}^{2}}}$
$45.9(2-3x+{{x}^{2}})\text{ = 9+4}{{\text{x}}^{2}}+12x$ by solving it we will get,
$82.8-149.7x+41.9{{\text{x}}^{2}}\text{ = 0}$
Now, by using the quadratic equation we will get:
$x\text{ = }\dfrac{-\text{b }\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
$x\text{ = }\dfrac{-149.7\text{ }\pm \sqrt{{{(-149.7)}^{2}}-4(41.9)(82.8)}}{2(41.9)}$
$x\text{ = 2}\text{.888 or 0}\text{.684}$
-So, the value of x = 2.888 is not taken as the value of concentration will become negative. So, we will use x = 0.684.
-Hence, the equilibrium concentration will be:
$({{\text{H}}_{2}})\ \text{= 1-x = 1-0}\text{.684 = 0}\text{.316M}$
$({{\text{I}}_{2}})\ \text{= 2-x = 2-0}\text{.684 = 1}\text{.316M}$
$(\text{HI})\ \text{= 3 + 2x = 3 + 2(0}\text{.684) = 4}\text{.36M}$
Note: Le- chatelier Principle states that when a reaction is in equilibrium state then any change in the reaction due to change in the conditions will affect the rate of equation. In the answer, the concentration of hydrogen, iodine was subtracted from x because the concentration reactant decreases as the process starts whereas 2x was added with the concentration of HI because the concentration of product increases.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Select the word that is correctly spelled a Twelveth class 10 english CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What organs are located on the left side of your body class 11 biology CBSE
What is BLO What is the full form of BLO class 8 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE