Answer:
[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=[tex]\frac{[H_{2}[I_{2} ] }{[HI]^{2}}[/tex] (1)
The initial concentration of HI is:
[HI] = [tex]\frac{0,0660}{2,00}[/tex] = 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 = [tex]\frac{[x]^2}{[0,0330-2x]^2}[/tex]
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!
Answer:
M (HI) = 0.033MHI
M (N2) = 0.0165MN2
M (H2) = 0.0165MH2
Explanation:
The Molarity formula is the amount of substance in moles (n) divided by one liter of solution (L)
M = n / L
M (HI) = n / V = 0.066mol HI / 2L solution
M (HI) = 0.033MHI
moles of H2 = 0.066molHI * (1mol H2 / 2 molesHI) = 0.033mol H2
moles of
M (H2) = n / V = 0.033molH2 / 2Ldisolution
M (H2) = 0.0165MH2
N2 = 0.066molHI * (1mol N2 / 2 molesHI) = 0.033mol N2
M (N2) = n / V = 0.033molN2 / 2L solution
M (N2) = 0.0165MN2
