Question

One mole of ${{\text{H}}_{2}}$, two moles of${{\text{I}}_{2}}$ and three moles of $\text{HI}$ are injected in a 1 litre flask. What will be the concentration of ${{\text{H}}_{2}},\ {{\text{I}}_{2}}\text{ and HI}$ at equilibrium at $\text{490}{}^\circ \text{C}$? The equilibrium constant for the reaction at $\text{490}{}^\circ \text{C}$ is 45.9.

Answer 1

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.