Question

when 0.80 moles of iodine and 0.80 moles of hydrogen are heated in a closed vessel at 444C , 0.60 moles of hydroiodic acid are formed at equilibrium the equilibrium constant kc of this reaction is ​

Answer 1

The equilibrium constant $K_c$ of this reaction is ​1.44

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_c$

For the given chemical reaction:

$H_2(g)+I_2(g)\rightarrow 2HI(g)$                           

Initial amount     0.80     0.80                0      

At eqm.             (0.80-x)   (0.80-x)         2x    

The expression for $K_c$ is written as:

$K_c=\frac{[HI]^2}{[H_2]^1[I_2]^1}$

$K_c=\frac{(2x)^2}{(0.80-x)^1\times (0.80-x)^1}$

Given : 2x= 0.60

x = 0.30

$K_c=\frac{(0.60)^2}{(0.80-0.30)^1\times (0.80-0.30)^1}$

$K_c=1.44$

The equilibrium constant $K_c$ of this reaction is ​1.44

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