For the following reaction, the equilibrium constant $K_{e}$ at $298 K$ is $1.6 \times 10^{17} .$
$F e^{2 +} (a q) + S^{2 -} (a q) ⇌ F e S (s)$
When equal volumes of $0.06 M F e^{2 +} (a q)$ and $0.2 M S^{2 -} (a q)$ solutions are mixed, the equilibrium concentration of $F e^{2 +} (a q)$ is found to be $Y \times 10^{- 17} M .$ The value of $Y$ is __________

Question

For the following reaction, the equilibrium constant

K_{e}

at

298 K

is

1.6 \times 10^{17} .

F e^{2 +} (a q) + S^{2 -} (a q) ⇌ F e S (s)

When equal volumes of

0.06 M F e^{2 +} (a q)

and

0.2 M S^{2 -} (a q)

solutions are mixed, the equilibrium concentration of

F e^{2 +} (a q)

is found to be

Y \times 10^{- 17} M .

The value of

Y

is __________

Aicharya · Accepted Answer

Fe 2+ + S 2- &#8652; Fe&#8201; S ( s ) ..... K c = 1.6 &#215; 10 17 t = 0 : 0.03 &#8201;M&#160; 0.1 &#8201;M t = t; &#160; 0.03 - H &#160; 0.1 &#8201;H Since K c &#62; &#62; 1 , n &#8771; 0.03 &#8756; &#8201; [ S 2- ] = 0.1 - 0.031 = 0.07 &#8201;M K c = Fe 2 + S 2- = 1 1.6 &#215; 10 17 Fe 2 + = 1 1.6 &#215; 0.07 &#215; 10 - 17 y = 1 1.6 &#215; 0.07 &#8771; 8.93

For the following reaction, the equilibrium constant K e at 298   K is 1.6 × 10 17 . F e 2 + a q +…

Solution

For the following reaction, the equilibrium constant K e at 298 &#160; K is 1.6 &#215; 10 17 . F e 2 + a q +…

Solution

For the following reaction, the equilibrium constant K e at 298 K is 1.6 × 10 17 . F e 2 + a q +…