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Question

Answers

A.$0.67$

B.$1.5$

C.$2.67\times {{10}^{4}}$

D.$9.0\times {{10}^{4}}$

Answer
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As we can express equilibrium constant of the reaction in terms of molar concentration of the reactant and product which we denote it as ${{K}_{C}}$.But for reaction involving gases which is given in this question it is more convenient to express the equilibrium constant in terms of partial pressure which we are denoting it as ${{K}_{P}}$.Sometimes these both equilibrium are same but not always.

To find the${{K}_{P}}$for the equilibrium ${{I}_{2}}(g)2I(g)$

Let us compare it with the general reaction $aA + bB and cC + dD$

For which ${{K}_{P}}=\dfrac{(p_{C}^{c})(p_{D}^{d})}{(p_{A}^{a})(p_{B}^{b})}$

${{K}_{P}}$$=\dfrac{{{[C]}^{c}}{{[D]}^{d}}{{(RT)}^{(c+d)}}}{{{[A]}^{a}}{{[B]}^{b}}{{(RT)}^{(a+b)}}}$

So no according to given equilibrium we can say that${{K}_{P}}=\dfrac{{{(p_{I}^{{}})}^{2}}}{(p_{{{I}_{2}}}^{{}})}$

The partial pressure of iodine atom will be:

$({{P}_{I}})=\dfrac{{{10}^{5}}\times 40}{100}$

$({{P}_{I}})=0.4\times {{10}^{6}}Pa$

Now the partial pressure of ${{I}_{2}}$ molecule will be:

$({{P}_{{{I}_{2}}}})=\dfrac{{{10}^{5}}\times 60}{100}$

$({{P}_{{{I}_{2}}}})=0.6\times {{10}^{5}}Pa$

Now putting values in the equation i.e.

${{K}_{P}}=\dfrac{{{(p_{I}^{{}})}^{2}}}{(p_{{{I}_{2}}}^{{}})}$

${{K}_{P}}=\dfrac{{{(0.4\times {{10}^{6}}Pa)}^{2}}}{0.6\times {{10}^{5}}Pa}$

Here we get the answer as ,

${{K}_{P}}=2.67\times {{10}^{4}}$

It must be remembered that for the existence of heterogeneous equilibrium pure solid and liquids must also be present at equilibrium, but their concentration or partial pressure do not appear in the expression of the equilibrium constant.

1$pascal$, $Pa=1N{{m}^{-2}}$and $1bar={{10}^{5}}Pa$.

Equilibrium constant can also be expressed as dimensionless quantities if the standard state of reactants and products are specific. Like for example a pure gas has a standard state as 1 bar.