A mixture of 1 mole of $H_{2} O$ and 1 mole of CO is taken in a 10 litre container and heated to 725 K. At equilibrium 40% of water by mass reacts with carbon monoxide according to the equation.
$C O (g) + H_{2} O_{(v a p)} \Leftrightarrow C O_{2} (g) + H_{2} (g)$
The equilibrium constant $K_{c} \times 10^{2}$ for the reaction is Q, Find Q/11 is______

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 4.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

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

Take Free Test

Detailed Solution

$\begin{array}{l} C O (g) + H_{2} O (g) \Leftrightarrow C O_{2} (g) + H_{2} (g) \\ t = 0; 1 m o l 1 m o l 0 0 \\ a t e q u .; 1 - x 1 - x x x \end{array}$

at equilibrium 40% by mass water reacts with CO

X=0.4 1-X=0.6

$K_{C} = \frac{[C O_{2}] [H_{2}]}{[C O] [H_{2} O]} = \frac{0.4 \times 0.4}{0.6 \times 0.6} = 0.44$

$K_{C} \times 10^{2} = 44;$ $Q = \frac{44}{11} = 4$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!