An equilibrium system for the reaction between hydrogen and iodine to give hydrogen iodide at 765 K in a 5 litre volume contains 0.4 mole of hydrogen, 0.4 mole of iodine and 2.4 moles of hydrogen iodide. The equilibrium constant for the reaction H2 + I2 leftharpoons 2HI is

Question

An equilibrium system for the reaction between hydrogen and iodine to give hydrogen iodide at 765 K in a 5 litre volume contains 0.4 mole of hydrogen, 0.4 mole of iodine and 2.4 moles of hydrogen iodide. The equilibrium constant for the reaction H2 + I2 leftharpoons 2HI is (A) 36.0 (B) 15.0 (C) 0.067 (D) 0.028

Tardigrade Admin · Accepted Answer

H2 + I2 leftharpoons 2HI At equilibrium (0.4/5) (0.4/5) (2.4/5) Kc=(((2.4/5))2/((0.4)5)((0.4)5) K/c) = (2.4×2.4 /0.4×0.4) Kc =(576/0.16) Kc = 36

Q. An equilibrium system for the reaction between hydrogen and iodine to give hydrogen iodide at $765 K$ in a $5$ litre volume contains $0.4$ mole of hydrogen, $0.4$ mole of iodine and $2.4$ moles of hydrogen iodide. The equilibrium constant for the reaction
$H_{2} + I_{2} ⇌ 2 H I$ is

$36.0$

$15.0$

$0.067$

$0.028$

$H_{2} + I_{2} ⇌ 2 H I$
At equilibrium $\frac{0.4}{5} \frac{0.4}{5} \frac{2.4}{5}$
$K_{c} = \frac{( \frac{2.4}{5} ) ^{2}}{( \frac{0.4}{5} ) ( \frac{0.4}{5} )}$
$K_{c} = \frac{2.4 \times 2.4}{0.4 \times 0.4}$
$K_{c} = \frac{576}{0.16}$
$K_{c} = 36$

Q. An equilibrium system for the reaction between hydrogen and iodine to give hydrogen iodide at 765K in a 5 litre volume contains 0.4 mole of hydrogen, 0.4 mole of iodine and 2.4 moles of hydrogen iodide. The equilibrium constant for the reaction H2​+I2​⇌2HI is

36.0

15.0

0.067

0.028