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For the formation of hydrogen iodide, H2(g) + 12(g) - 2HI(g) the value of the
equilibrium constant at 700K is 54.7. If one mole of H2 and one mole of 12 are the
only materials initially present, what will be the equilibrium concentrations of H2,
12, and HI.

Respuesta :

tcs1030318

Answer:

t 700 K, the equilibrium constant for the reaction;

H  

2(g)

​  

+I  

2(g)

​  

⇌2HI  

(g)

​  

 is 54.8.

If 0.5 mol litre  

−1

 of HI  

(g)

​  

 is present at equilibrium at 700 K, what are the concentrations of H  

2(g)

​  

 and I  

2(g)

​  

 assuming that we initially started with HI  

(g)

​  

 and allowed it to reach equilibrium at 700 K.

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