Question

At 490°C, the equilibrium constant for the synthesis of HI is
50, the value of K for the dissociation of HI will be
(a) 20.0
(b) 2.0
(c) 0.2
(d) 0.02​

Answer 1

Answer:

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Answer 2

The value of K for the dissociation of HI will be 0.02

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

$\frac{1}{2}H_2+\frac{1}{2}I_2\rightarrow HI$

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

$K=50$

On reversing the reaction:

$HI\rightarrow \frac{1}{2}H_2+\frac{1}{2}I_2$

$K'=\frac{[H_2]^\frac{1}{2}\times [I_2]^\frac{1}{2}}{HI]}$

$K'=\frac{1}{K}=\frac{1}{50}=0.02$

Thus the value of K for the dissociation of HI will be 0.02

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