Question

The following reaction was performed in a sealed vessel at 772 ∘C : H2(g)+I2(g)⇌2HI(g) Initially, only H2 and I2 were present at concentrations of [H2]=3.90M and [I2]=2.95M . The equilibrium concentration of I2 is 0.0500 M . What is the equilibrium constant, Kc, for the reaction at this temperature?

Answer 1

Answer: The value of $K_c$ for the given equation is 160.2

Explanation:

We are given:

Initial concentration of hydrogen gas = 3.90 M

Initial concentration of iodine gas = 2.95 M

Equilibrium concentration of iodine gas = 0.0500 M

The given chemical equation follows:

                    $H_2(g)+I_2(g)\rightleftharpoons 2HI(g)$

Initial:          3.90     2.95

At eqllm:   3.90-x    2.95-x      2x

Evaluating the value of 'x':

$\Rightarrow (2.95-x)=0.0500\\\\x=2.9$

So, equilibrium concentration of hydrogen gas = (3.95 - x) = (3.95 - 2.9) = 1.05 M

Equilibrium concentration of HI gas = x = 2.9 M

The expression of $K_c$ for above equation follows:

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

Putting values in above equation, we get:

$K_c=\frac{(2.9)^2}{1.05\times 0.0500}\\\\K_c=160.2$

Hence, the value of $K_c$ for the given equation is 160.2