Question
Observe the following equilibrium in a 1 L flask.
$$A(g) \rightleftharpoons B(g)$$
At $T(K)$, the equilibrium concentrations of A and B are 0.5 M and 0.375 M respectively. 0.1 moles of A is added into the flask and heated to $T(K)$ to establish the equilibrium again. The new equilibrium concentrations (in M ) of A and B are respectively.
Options
Solution
$A \rightleftharpoons B$
$0.5M\ 0.375M\ $ (At equilibrium)
$$K_{eq} = \frac{\lbrack B\rbrack_{eq}}{\lbrack A\rbrack_{eq}} = \frac{0.375}{0.5} = 0.75 $$Now 0.1 mole of A is added so reaction will move in forward direction.
$$A \rightleftharpoons \text{ }B$$
$${0.6 - x\ 0.375 + x }{K_{eq} = 0.75 = \frac{0.375 + x}{0.6 - x} }{0.45 - 0.75x = 0.375 + x }{1.75x = 0.075 }{X = \frac{0.075}{1.75} = \frac{3}{70} = 0.043 }$$Moles of $A = 0.043 = 0.557$
Moles of $B = 0.418$
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