When 2.00mol each of $H_{2} (g)$ and $I_{2} (g)$ are reacted in a 1.00 L container at a certain temperature. 3.50mol of HI is present at equilibrium. Calculate the value of the equilibrium constant, $K_{c}$ for the reaction:
$H_{2} + I_{2} ⇌ 2 HI$

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$2.0 \times 10^{2}$

3.7

answer is D.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

The reaction can be given as,

$\underset{2 - x}{H_{2}} + \underset{2 - x}{I_{2}} ⇌ \underset{2 x = 3.50}{2 HI}$

For this the equilibrium constant,

$K_{C} = \frac{(3.5)^{2}}{(0.25)^{2}} K_{C} = 196$

So, the value of the equilibrium constant, $K_{c}$ for the reaction is $2.0 \times 10^{2}$ .

Watch 3-min video & get full concept clarity

No courses found

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test