Question

0.5 mole of H2(g) and 1 mole of HI (g) (but no I2) are added to a 1lt vessel and allowed to reach equilibrium according to the following reaction: H2+I2=2HI. If x is the equilibrium concentration of I2 the correct equilibrium constant expression is:(1) X(0.5-x)/(1+2x)^2. (2) (2-x)^2/(0.5+X)^2. (3) X(2+X)^2/(0.5-x). (4) (1-2x)^2/(0.5+X)X.​

Answer 1

Answer: The correct answer is Option 4.

Explanation:

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric coefficients. It is represented by $K_{c}$

For a general chemical reaction:

$aA+bB\rightleftharpoons cC+dD$

The $K_{c}$ is written as:

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

We are given:

Initial concentration of $H_2=\frac{0.5}{1}=0.5$

Initial concentration of $HI=\frac{1}{1}=1$

For the given chemical reaction:

                             $H_2+I_2\rightleftharpoons 2HI$

Initial:                   0.5              1

At eqllm:          (0.5+x)     x    (1-2x)

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[HI]^2}{[H_2][I_2]}$

$K_c=\frac{(1-2x)^2}{(0.5+x)\times x}$

Hence, the correct answer is Option 4.

Answer 2

Answer:

(4)

here's the answer and hope this helps you.