At a certain temperature the equilibrium constant KC is 0.25 for the reaction A(.g.)+B(.g.)leftharpoons C(.g.)+D(.g.) If we take 1 mole of each of four gases in a 10 litre container, what would be the equilibrium concentration of A(g) ?

Question

At a certain temperature the equilibrium constant KC is 0.25 for the reaction A(.g.)+B(.g.)leftharpoons C(.g.)+D(.g.) If we take 1 mole of each of four gases in a 10 litre container, what would be the equilibrium concentration of A(g) ? (A) 0.331 M (B) 0.033 M (C) 0.133 M (D) 1.33 M

Tardigrade Admin · Accepted Answer

The give reaction will go in backward direction

Solution

K_{c} = \frac{( A ) \times ( B )}{( C ) \times ( D )}

4 = \frac{( 1 + α ) ^{2}}{( 1 - α ) ^{2}}

2 = (\frac{1 + α}{1 - α})

2 - 2 α = 1 + α

α = \frac{1}{3} = 0.33

[A] = \frac{1 + α}{10}

= \frac{1 + 0.33}{10} = 0.133

M

Q. At a certain temperature the equilibrium constant $K_{C}$ is $0.25$ for the reaction

$A (g) + B (g) ⇌ C (g) + D (g)$

If we take $1$ mole of each of four gases in a $10$ litre container, what would be the equilibrium concentration of $A (g)$ ?

0.331 M

0.033 M

0.133 M

1.33 M

The give reaction will go in backward direction

$K_{c} = \frac{( A ) \times ( B )}{( C ) \times ( D )}$

$4 = \frac{( 1 + α ) ^{2}}{( 1 - α ) ^{2}}$

$2 = (\frac{1 + α}{1 - α})$

$2 - 2 α = 1 + α$

$α = \frac{1}{3} = 0.33$

$[A] = \frac{1 + α}{10}$

$= \frac{1 + 0.33}{10} = 0.133$ M

Q. At a certain temperature the equilibrium constant KC​ is 0.25 for the reaction A(g)+B(g)⇌C(g)+D(g) If we take 1 mole of each of four gases in a 10 litre container, what would be the equilibrium concentration of A(g) ?