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Chemistry

This is a Multiple part question. Please write step by step thatway I can study from this problem. Thank you.
(a).
Complete the following equilibrium table.
2HI_(g)Â Â Â Â Â Â â‡ŒÂ Â H_2(g) + I_2(g)
InitialÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.1600 MÂ Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â 0
ChangeÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€
EquilibriumÂ Â Â Â
(b). Determine equilibrium values for [HI] and [H₂]if [I₂] = 0.0227 M at equilibrium.
(c). Write the expression (with substances in brackets) forK_C.
(d). Substitute equilibrium concentrations in the expression,and determine the value for K_C.
(e). Write the expression (with substances in brackets) forQ_C.Include the â€œiâ€ subscripts.
What does the â€œiâ€ stand for? Describe the difference betweenQ_C and Kc.

Answer & Explanation Solved by verified expert

4.3 Ratings (572 Votes)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 2HI_(g)Â Â Â Â Â Â â‡Œ Â Â H_2(g) + I_2(g)

IÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.16Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â 0

CÂ Â Â Â Â Â Â Â Â Â Â Â Â -2*0.0227Â Â Â Â Â Â Â 0.0227 0.0227

EÂ Â Â Â Â Â Â Â Â Â Â 0.16-2*0.0227 0.0227Â Â 0.0227

Â Â Â Â Â Â Â Â Â Â Â Â Â 0.1146

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Kc = [H2][I2]/[HI]²

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.0227*0.0227/(0.114)²Â Â Â = 0.039

Kc>Qc the equilibrium shift ot right side.

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