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JEE MAIN - Chemistry (2022 - 24th June Morning Shift - No. 4)

For a reaction at equilibrium

A(g) $$\rightleftharpoons$$ B(g) + $${1 \over 2}$$ C(g)

the relation between dissociation constant (K), degree of dissociation ($$\alpha$$) and equilibrium pressure (p) is given by :

$$K = {{{\alpha ^{{1 \over 2}}}{p^{{3 \over 2}}}} \over {{{\left( {1 + {3 \over 2}\alpha } \right)}^{{1 \over 2}}}(1 - \alpha )}}$$
$$K = {{{\alpha ^{{3 \over 2}}}{p^{{1 \over 2}}}} \over {{{\left( {2 + \alpha } \right)}^{{1 \over 2}}}(1 - \alpha )}}$$
$$K = {{{{(\alpha \,p)}^{{3 \over 2}}}} \over {{{\left( {1 + {3 \over 2}\alpha } \right)}^{{1 \over 2}}}(1 - \alpha )}}$$
$$K = {{{{(\alpha \,p)}^{{3 \over 2}}}} \over {{{\left( {1 + \alpha } \right)}}{{(1 - \alpha )}^{{1 \over 2}}}}}$$

Explanation

JEE Main 2022 (Online) 24th June Morning Shift Chemistry - Chemical Equilibrium Question 33 English Explanation

Now,

$${K_P}$$ or $$K = {{{P_B} \times {{\left( {{P_C}} \right)}^{{1 \over 2}}}} \over {{P_A}}}$$

$$ = {{\left( {{\alpha \over {1 + {\alpha \over 2}}}} \right)P \times {{\left[ {\left( {{{{\alpha \over 2}} \over {1 + {\alpha \over 2}}}} \right)P} \right]}^{{1 \over 2}}}} \over {\left( {{{1 - \alpha } \over {1 + {\alpha \over 2}}}} \right)P}}$$

$$ = {{\left( {{{2\alpha } \over {2 + \alpha }}} \right)P \times {{\left[ {\left( {{\alpha \over {2 + \alpha }}} \right)P} \right]}^{{1 \over 2}}}} \over {\left( {{{2(1 - \alpha )} \over {2 + \alpha }}} \right)P}}$$

$$= {\alpha \over {1 - \alpha }} \times {\left( {{{\alpha P} \over {2 + \alpha }}} \right)^{{1 \over 2}}}$$

$$ = {{{\alpha ^{{3 \over 2}}}\,.\,{P^{{1 \over 2}}}} \over {(1 - \alpha ){{(2 + \alpha )}^{{1 \over 2}}}}}$$

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