Kc = 1.88 for 2HI(g) ↔ H2(g) + I2(g)

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SUMMARY

The equilibrium constant Kc for the reaction H2(g) + I2(g) ↔ 2HI(g) is calculated using the known Kc value of 1.37 for the reaction HI(g) ↔ 1/2 H2(g) + 1/2 I2(g). By reversing and doubling the reaction, the relationship between Kc values can be established. The resulting Kc value for the desired reaction is Kc = 1.88, derived from the mathematical manipulation of the initial Kc value.

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Madelin Pierce
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Homework Statement


Kc = 1.37 for HI(g) ↔ 1/2 H2(g) + 1/2 I2(g). What is Kc (to 3 decimal places) for H2(g) + I2(g) ↔ 2HI(g)?

Homework Equations


Kc= Products/Reactants, Ice Tables

The Attempt at a Solution


I tried an ice table, but I’m not sure what to do with the Kc value I’m given when I have to another Kc value
 
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$$K_{c_1} = 1.37 = \dfrac{[H_2]^{1/2} [I_2]^{1/2}}{[HI]}$$ and $$K_{c_2} = \dfrac{[HI]^2}{[H_2][I_2]}$$ Can you find a relation between ##K_{c_1}## and ##K_{c_2}##?
 
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I know the reaction reversed and doubled. But I don’t know how to solve for Kc mathematically with this set-up.
 
Madelin Pierce said:
But I don’t know how to solve for Kc mathematically with this set-up.
What maths are you referring to? If you can find ##K_{c_2}## in terms of ##K_{c_1}##, then can't you find ##K_{c_2}##?
 
Madelin Pierce said:
I know the reaction reversed and doubled. But I don’t know how to solve for Kc mathematically with this set-up.
What do you get if you square @Wrichik Basu's first equation, and then multiply it by his second equation.
 

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