text15 moles of H2 and 5.2 moles of I2 are mixed and allowed to attain equilibrium at 500° C . At equilibrium, the number of moles of HI is found to be 10 mole. The equilibrium constant for the formation of HI is

Question

text15 moles of H2 and 5.2 moles of I2 are mixed and allowed to attain equilibrium at 500° C . At equilibrium, the number of moles of HI is found to be 10 mole. The equilibrium constant for the formation of HI is (A) 50 (B) 15 (C) 100 (D) 25

Tardigrade Admin · Accepted Answer

H2 + I2 leftharpoons 2HI 15 5.2 0 (15 - 5) (5.2 - 5) 10 Equilibrium constant (( textK) textc)=(([H I])2/[H2] [. I2 ].)=(10 × 10/10 × 0.2)=50

Q. $15$ moles of $H_{2}$ and $5.2$ moles of $I_{2}$ are mixed and allowed to attain equilibrium at $50 0^{\circ} C$ . At equilibrium, the number of moles of $H I$ is found to be $10$ mole. The equilibrium constant for the formation of $H I$ is

$50$

$15$

$100$

$25$

$H_{2} + I_{2} ⇌ 2 H I$
$15$ $5.2$ $0$
$(15 - 5)$ $(5.2 - 5)$ $10$
Equilibrium constant
$((K)_{c}) = \frac{( [ H I ] ) ^{2}}{[ H _{2} ] [ I _{2} ]} = \frac{10 \times 10}{10 \times 0.2} = 50$

Q. 15 moles of H2​ and 5.2 moles of I2​ are mixed and allowed to attain equilibrium at 500∘C . At equilibrium, the number of moles of HI is found to be 10 mole. The equilibrium constant for the formation of HI is

50

15

100

25