Answer: Equilibrium constant is 0.70.
Explanation:
Initial moles of
= 0.35 mole
Volume of container = 1 L
Initial concentration of
Initial moles of
= 0.40 mole
Volume of container = 1 L
Initial concentration of
equilibrium concentration of
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The given balanced equilibrium reaction is,

Initial conc. 0.35 M 0.40M 0 0
At eqm. conc. (0.35-x) M (0.40-x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[CO_2]\times [H_2O]}{[CO]\times [H_2O]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO_2%5D%5Ctimes%20%5BH_2O%5D%7D%7B%5BCO%5D%5Ctimes%20%5BH_2O%5D%7D)

we are given : (0.35-x)= 0.18
x = 0.17
Now put all the given values in this expression, we get :


Thus the value of the equilibrium constant is 0.70.