Equilibrium constant for the given reaction is K =1020 at temperature 300 K A ( s )+2 B text (aq.) leftharpoons 2 C text (s) + D text (aq.) K =1020 Calculate the equilibrium conc. of B starting with mixture of 1 mole of A and 1 / 4 mole/litre of B at 300 K (Give your answer by multiplying it by 1011 )

Question

Equilibrium constant for the given reaction is K =1020 at temperature 300 K A ( s )+2 B text (aq.) leftharpoons 2 C text (s) + D text (aq.) K =1020 Calculate the equilibrium conc. of B starting with mixture of 1 mole of A and 1 / 4 mole/litre of B at 300 K (Give your answer by multiplying it by 1011 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/a068de3f7d2ba585476e046c0b1e9d2e-.png" /> 1020=((1/4)/[ B ]2) ⇒ 1020=((1/4)/a2) ⇒ a2=(1/4 × 1020)=(10-20/4) ⇒ a=(10-10/2)=5 × 10-11 M

Q. Equilibrium constant for the given reaction is K=1020 at temperature 300K A(s)+2B (aq.) ⇌2C (s) +D (aq.) K=1020 Calculate the equilibrium conc. of B starting with mixture of 1 mole of A and 1/4 mole/litre of B at 300K (Give your answer by multiplying it by 1011 )

1020=[B]241​​⇒1020=a241​​ ⇒a2=4×10201​=410−20​ ⇒a=210−10​=5×10−11M

$1 0^{20} = \frac{\frac{1}{4}}{[ B ] ^{2}} \Rightarrow 1 0^{20} = \frac{\frac{1}{4}}{a ^{2}}$
$\Rightarrow a^{2} = \frac{1}{4 \times 1 0 ^{20}} = \frac{1 0 ^{- 20}}{4}$
$\Rightarrow a = \frac{1 0 ^{- 10}}{2} = 5 \times 1 0^{- 11} M$