Question

1.50 mol each of hydrogen and iodine were placed in a sealed 10 L container maintained at 717 K. At equilibrium 1.25 mol each of hydrogen and iodine were left behind. The equilibrium constant for the reaction H2 + I2 ->2HI at 717 K is

Answer 1

Answer: Equilibrium constant for the reaction is 0.16.

Explanation: We are given:

Moles of Hydrogen = 1.50 moles

Moles of Iodine = 1.50 moles

Volume = 10 L

Concentration or Molarity is calculated by:

$Molarity=\frac{Moles}{Volume}$

$\text{Molarity of Hydrogen (Initially)}=\frac{1.50mol}{10L}=0.15M$

$\text{Molarity of Iodine (Initially)}=\frac{1.50mol}{10L}=0.15M$

$\text{Molarity of Hydrogen (At equilibrium)}=\frac{1.25mol}{10L}=0.125M$

$\text{Molarity of Iodine (At equilibrium)}=\frac{1.25mol}{10L}=0.125M$

We are given a reaction of the formation of hydrogen iodide, reaction follows:

                  $H_2(g)+I_2(g)\rightarrow 2HI((aq.)$

at $t=0$      0.15M        0.15M         0

at $t=t_{eq}$      (0.15-x)      (0.15-x)    2x

Concentration of hydrogen at equilibrium = 0.125M

0.15 - x = 0.125

x = 0.025 M

Concentration of Hydrogen iodide at equilibrium = 2x = 2(0.025) = 0.05M

Equilibrium constant for the reaction is given as:

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

Putting the values in above equation, we get

$K_c=\frac{(0.05)^2}{(0.125)(0.125)}=0.16$

Answer 2

Answer:

0.16

Explanation: