Hydrogen iodide decomposes according to the equation:
2hi(g)h2(g)
+ i2(g),kc = 0.0156 at 400oc
a 0.0660 mole sample of hi was injected into a 2.00 l
reactionvessel held at 400oc.
calculate the concentration of hi and i2
andh2 at equilibrium.

[HI] = 0,0264 M

[H₂] = 0,003299 M

[I₂] = 0,003299 M

Explanation:

For the equation:

2 HI (g) ⇄ H₂(g) + I₂(g)  With k= 0,0156. Also, k=(1)

The initial concentration of HI is:

[HI] = = 0,0330 M

The concentration of the reactant and products in equilibrium is:

[HI] = 0,033 M - 2x ⇒ because 2 moles are consumed

[H₂] = x ⇒ because 1 mole is produced

[I₂] = x ⇒ Also, 1 mole is produced

Thus, replacing in (1):

0.0156 =

The quadratic equation to solve is:

0,9376x² + 0,0020592x - 0,000017 = 0

Have two solutions:

x = -0,0054955  ⇒ No physical sense. Will produce negative concentrations

x = 0,003299⇒ Real answer

Thus, concentrations in equilibrium are:

[HI] = 0,033 - 2×0,003299 = 0,0264 M

[H₂] = 0,003299 M

[I₂] = 0,003299 M

I hope it helps!

8 carbon atoms forming single bonds with 18 hydrogen atoms; a triple bond exists between the third and fourth carbon atom from the right

explanation:

petroleum? coal maybe?