Question

for areversible reaction 2A+B→←3C+3D. write the expressionfor the equilibrium constant. ​

Answer 1

Answer:

Equilibrium constant for the given equation = $\dfrac{[C]^3[D]^3}{[A]^2[B]}$

Explanation:

$\blacksquare$ Reaction quotient

$\implies$Is the measure of relative amount of product and reactant

$\implies$It is denoted by the letter  $Q$

$\blacksquare$ Equilibrium constant

$\implies$Is the value of reaction quotient when a reversible eaction attains state of equilibrium ie, rate of forward reaction = rate of backward reaction

$\implies$It is denoted by $K_c$

For a given reaction,

aA + bB ${\rightleftharpoons}$ cC + dD

The reaction quotient,

$Q = \dfrac{[C]^c[D]^d}{[A]^a[B]^b}$

At equilibrium, the equilibrium constant

$K_c = \dfrac{[C]^c[D]^d}{[A]^a[B]^b}$

Hence, equilibrium constant for the given reaction

2A + B ${\rightleftharpoons}$ 3C + 3D is given by

$K_c =$  $\dfrac{[C]^3[D]^3}{[A]^2[B]}$