Fe3+(aq)+SCN−(aq)⇌FeSCN2+(aq)
A solution is made containing initial [Fe3+]= 1.5×10−3 M and initial [SCN−]= 6.5×10−4 M . At equilibrium, [FeSCN2+]= 1.7×10−4 M .
Finding Equilibrium Constant...Keq=?
2018-07-25 11:58 am
回答 (1)
2018-07-25 12:13 pm
Initial [FeSCN²⁺]ₒ = 0 M
[FeSCN²⁺] at equilibrium = 1.7 × 10⁻⁴ M
Increase in [FeSCN²⁺] = 1.7 × 10⁻⁴ M
_______________ Fe³⁺(aq) ____ + ____ SCN⁻(aq) ____ ⇌ ____ FeSCN²⁺(aq) ____ Keq
Initial: _______ 1.5 × 10⁻³ M ______ 6.5 × 10⁻⁴ M ___________ 0 M
Change: _____ -1.7 × 10⁻⁴ M _____ -1.7 × 10⁻⁴ M _________ +1.7 × 10⁻⁴ M
Equilibrium: _ 1.33 × 10⁻³ M ____ 4.8 × 10⁻⁴ M _________ 1.7 × 10⁻⁴ M
Keq = [FeSCN²⁺] / ([Fe³⁺] [SCN⁻]) = (1.7 × 10⁻⁴) / {(1.33 × 10⁻³) × (4.8 × 10⁻⁴)} = 266.3 ≈ 270 (to 2 sig. fig.)
[FeSCN²⁺] at equilibrium = 1.7 × 10⁻⁴ M
Increase in [FeSCN²⁺] = 1.7 × 10⁻⁴ M
_______________ Fe³⁺(aq) ____ + ____ SCN⁻(aq) ____ ⇌ ____ FeSCN²⁺(aq) ____ Keq
Initial: _______ 1.5 × 10⁻³ M ______ 6.5 × 10⁻⁴ M ___________ 0 M
Change: _____ -1.7 × 10⁻⁴ M _____ -1.7 × 10⁻⁴ M _________ +1.7 × 10⁻⁴ M
Equilibrium: _ 1.33 × 10⁻³ M ____ 4.8 × 10⁻⁴ M _________ 1.7 × 10⁻⁴ M
Keq = [FeSCN²⁺] / ([Fe³⁺] [SCN⁻]) = (1.7 × 10⁻⁴) / {(1.33 × 10⁻³) × (4.8 × 10⁻⁴)} = 266.3 ≈ 270 (to 2 sig. fig.)
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https://hk.answers.yahoo.com/question/index?qid=20180725035815AAGueh4