CHEM 120 Equilibrium Review Problem Set

Equilibrium Problem

  • Initial moles: H2 (5.443), Cl2 (7.432), HCl (3.211) in 1.500 L container.
  • Equilibrium constant: Kc = 30.5
  • Reaction: H<em>2(g)+Cl</em>2(g)⇌2HCl(g)H<em>2(g) + Cl</em>2(g) \rightleftharpoons 2HCl(g)
  • Initial concentrations: [H<em>2]=3.629M[H<em>2] = 3.629 M, [Cl</em>2]=4.955M[Cl</em>2] = 4.955 M, [HCl]=2.141M[HCl] = 2.141 M

Reaction Quotient and Direction

  • Q=[HCl]2[H<em>2][Cl</em>2]=(2.141)2(3.629)(4.955)=0.2549Q = \frac{[HCl]^2}{[H<em>2][Cl</em>2]} = \frac{(2.141)^2}{(3.629)(4.955)} = 0.2549
  • Q < K, reaction proceeds to the right (toward products).

ICE Table

  • R: H<em>2H<em>2, Cl</em>2Cl</em>2, 2HCl2HCl
  • I: 3.629, 4.955, 2.141
  • C: -x, -x, +2x
  • E: 3.629-x, 4.955-x, 2.141+2x

Equilibrium Expression

  • K=[HCl]2[H<em>2][Cl</em>2]K = \frac{[HCl]^2}{[H<em>2][Cl</em>2]}
  • 30.5=(2.141+2x)2(3.629−x)(4.955−x)30.5 = \frac{(2.141 + 2x)^2}{(3.629 - x)(4.955 - x)}
  • 30.5=4.584+8.564x+4x217.98−8.584x+x230.5 = \frac{4.584 + 8.564x + 4x^2}{17.98 - 8.584x + x^2}
  • 4.584+8.564x+4x2=548.39−261.84x+30.5x24.584 + 8.564x + 4x^2 = 548.39 - 261.84x + 30.5x^2
  • 26.5x2−270.40x+534.81=026.5x^2 - 270.40x + 534.81 = 0

Solving for x

  • Quadratic formula: x=−b±b2−4ac2ax = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}
  • x=−(−270.40)±(−270.40)2−4(26.5)(534.81)2(26.5)x = \frac{-(-270.40) ± \sqrt{(-270.40)^2 - 4(26.5)(534.81)}}{2(26.5)}
  • x=2.755x = 2.755 and x=7.448x = 7.448
  • x = 2.755 (7.448 is not possible due to negative concentration)

Equilibrium Concentrations

  • [H2]=3.629−2.755=0.874M[H_2] = 3.629 - 2.755 = 0.874 M
  • [Cl2]=4.955−2.755=2.200M[Cl_2] = 4.955 - 2.755 = 2.200 M
  • [HCl]=2.141+2(2.755)=7.651M[HCl] = 2.141 + 2(2.755) = 7.651 M

Checking the Work

  • Q=[HCl]2[H<em>2][Cl</em>2]=(7.651)2(0.874)(2.200)=30.444Q = \frac{[HCl]^2}{[H<em>2][Cl</em>2]} = \frac{(7.651)^2}{(0.874)(2.200)} = 30.444
  • Q ≈ K, solution is correct.

Partial Pressures at Equilibrium

  • Reaction: C(s)+O<em>2(g)⇌CO</em>2(g)C(s) + O<em>2(g) \rightleftharpoons CO</em>2(g)
  • ΔGrxno=−394.4kJmol\Delta G_{rxn}^o = -394.4 \frac{kJ}{mol}

Calculating Kp

  • ΔGo=−RTln(K)\Delta G^o = -RTln(K)
  • −394400Jmol=−(8.314Jmol⋅K)(298.15K)ln(K)-394400 \frac{J}{mol} = -(8.314 \frac{J}{mol \cdot K})(298.15 K)ln(K)
  • ln(K)=−394400Jmol−(8.314Jmol⋅K)(298.15K)=159.11ln(K) = \frac{-394400 \frac{J}{mol}}{-(8.314 \frac{J}{mol \cdot K})(298.15 K)} = 159.11
  • K=e159.11=1.26×1069=P<em>CO</em>2P<em>O</em>2K = e^{159.11} = 1.26 × 10^{69} = \frac{P<em>{CO</em>2}}{P<em>{O</em>2}}

Limiting Reactant

  • 100.0 g O<em>2×1molO</em>232.00gO<em>2=3.125molO</em>2O<em>2 × \frac{1 mol O</em>2}{32.00 g O<em>2} = 3.125 mol O</em>2
  • 100.0 g C×1molC12.01gC=8.326molCC × \frac{1 mol C}{12.01 g C} = 8.326 mol C
  • Oxygen is the limiting reactant.

Initial Partial Pressure of Oxygen

  • P<em>O</em>2=(3.125molO2)(0.08206L⋅atmmol⋅K)(298.15K)10.00L=7.646atmP<em>{O</em>2} = \frac{(3.125 mol O_2)(0.08206 \frac{L \cdot atm}{mol \cdot K})(298.15 K)}{10.00 L} = 7.646 atm
  • Initially, P<em>CO</em>2=0P<em>{CO</em>2} = 0

Equilibrium Partial Pressures

  • 1.26×1069=x7.646−x1.26 × 10^{69} = \frac{x}{7.646 - x}
  • 9.634×1069−1.26×1069x=x9.634 × 10^{69} - 1.26 × 10^{69}x = x
  • x=9.634×10691.26×1069=7.65atmCO2x = \frac{9.634 × 10^{69}}{1.26 × 10^{69}} = 7.65 atm CO_2

Solving for Partial Pressure of Oxygen

  • 1.26×1069=7.646x1.26 × 10^{69} = \frac{7.646}{x}
  • x=7.6461.26×1069=6.07×10−69atmO2x = \frac{7.646}{1.26 × 10^{69}} = 6.07 × 10^{-69} atm O_2

Checking the Work

  • Q=P<em>CO</em>2P<em>O</em>2=7.6466.07×10−69=1.26×1069Q = \frac{P<em>{CO</em>2}}{P<em>{O</em>2}} = \frac{7.646}{6.07 × 10^{-69}} = 1.26 × 10^{69}
  • Q = K, so answers are correct.

Weak Acid Equilibrium

  • Weak monoprotic acid HA with Ka=1.54×10−7K_a = 1.54 \times 10^{-7}
  • 0.10 M solution of HA.
  • HA(aq)+H<em>2O(l)⇌A−(aq)+H</em>3O+(aq)HA(aq) + H<em>2O(l) \rightleftharpoons A^-(aq) + H</em>3O^+(aq)

ICE Table

  • I: 0.10, 0, 0
  • C: -x, +x, +x
  • E: 0.10 - x, x, x

Approximation

  • Since Ka<10−3K_a < 10^{-3} and [HA]>0.05M[HA] > 0.05 M, neglect x in (0.10 - x).
  • K<em>a=[A−][H</em>3O+][HA]=1.54×10−7=(x)(x)0.10K<em>a = \frac{[A^-][H</em>3O^+]}{[HA]} = 1.54 \times 10^{-7} = \frac{(x)(x)}{0.10}
  • x2=1.54×10−8x^2 = 1.54 \times 10^{-8}
  • x=1.24×10−4=[A−]=[H3O+]x = 1.24 \times 10^{-4} = [A^-] = [H_3O^+]

pH Calculation

  • pH=−log([H3O+])=−log(1.24×10−4)=3.906pH = -log([H_3O^+]) = -log(1.24 \times 10^{-4}) = 3.906

Weak Base Equilibrium

  • Weak base B with Kb=3.84×10−5K_b = 3.84 \times 10^{-5}
  • 0.200 M solution of B.
  • B(aq)+H2O(l)⇌BH+(aq)+OH−(aq)B(aq) + H_2O(l) \rightleftharpoons BH^+(aq) + OH^-(aq)

ICE Table

  • I: 0.20, 0, 0
  • C: -x, +x, +x
  • E: 0.20 - x, x, x

Approximation

  • Since Kb<10−3K_b < 10^{-3} and [B]>0.05M[B] > 0.05 M, neglect x in (0.20 - x).
  • Kb=[BH+][OH−][B]=3.84×10−5=(x)(x)0.20K_b = \frac{[BH^+][OH^-]}{[B]} = 3.84 \times 10^{-5} = \frac{(x)(x)}{0.20}
  • x2=7.68×10−6x^2 = 7.68 \times 10^{-6}
  • x=2.77×10−3=[BH+]=[OH−]x = 2.77 \times 10^{-3} = [BH^+] = [OH^-]

pOH and pH Calculation

  • pOH=−log([OH−])=−log(2.77×10−3)=2.557pOH = -log([OH^-]) = -log(2.77 \times 10^{-3}) = 2.557
  • pH=14−pOH=14−2.557=11.443pH = 14 - pOH = 14 - 2.557 = 11.443

Solubility of Copper(II) Hydroxide

  • Saturated solution of Cu(OH)<em>2Cu(OH)<em>2, K</em>sp=2.2×10−20K</em>{sp} = 2.2 \times 10^{-20}
  • Cu(OH)2(s)⇌Cu2+(aq)+2OH−(aq)Cu(OH)_2(s) \rightleftharpoons Cu^{2+}(aq) + 2OH^-(aq)
  • Ksp=[Cu2+][OH−]2K_{sp} = [Cu^{2+}][OH^-]^2

Equilibrium Concentrations

  • [Cu2+]=x[Cu^{2+}] = x, [OH−]=2x[OH^-] = 2x
  • Ksp=(x)(2x)2=4x3K_{sp} = (x)(2x)^2 = 4x^3
  • 2.2×10−20=4x32.2 \times 10^{-20} = 4x^3
  • x3=5.5×10−21x^3 = 5.5 \times 10^{-21}
  • x=5.5×10−213=1.77×10−7Mx = \sqrt[3]{5.5 \times 10^{-21}} = 1.77 \times 10^{-7} M
  • [Cu2+]=1.77×10−7M[Cu^{2+}] = 1.77 \times 10^{-7} M, [OH−]=3.53×10−7M[OH^-] = 3.53 \times 10^{-7} M

Acetic Acid Buffer

  • Acetic acid buffer with pH = 5.00
  • Total acetic acid + acetate concentration = 100.0 mM
  • pKa=4.74pK_a = 4.74
  • CH<em>3COOH(aq)+H</em>2O(l)⇌CH<em>3COO−(aq)+H</em>3O+(aq)CH<em>3COOH(aq) + H</em>2O(l) \rightleftharpoons CH<em>3COO^-(aq) + H</em>3O^+(aq)

Buffer Equation

  • pH=pK<em>a+log([CH</em>3COO−][CH3COOH])pH = pK<em>a + log( \frac{[CH</em>3COO^-]}{[CH_3COOH]})
  • 5=4.74+log(x100−x)5 = 4.74 + log( \frac{x}{100 - x})
  • 0.26=log(x100−x)0.26 = log( \frac{x}{100 - x})
  • 100.26=x100−x10^{0.26} = \frac{x}{100 - x}
  • 1.8197=x100−x1.8197 = \frac{x}{100 - x}
  • 181.97−1.8197x=x181.97 - 1.8197x = x
  • 181.97=2.8197x181.97 = 2.8197x
  • x=181.972.8197=64.5mM[CH3COO−]x = \frac{181.97}{2.8197} = 64.5 mM [CH_3COO^-]
  • [CH3COOH]=100.0mM−64.5mM=35.5mM[CH_3COOH] = 100.0 mM - 64.5 mM = 35.5 mM

Molar Solubility of Calcium Fluoride

  • CaF<em>2(s)⇌Ca2+(aq)+2F−(aq)CaF<em>2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq), K</em>sp=3.9×10−11K</em>{sp} = 3.9 \times 10^{-11}
  • Ksp=[Ca2+][F−]2=(x)(2x)2=4x3K_{sp} = [Ca^{2+}][F^-]^2 = (x)(2x)^2 = 4x^3
  • 4x3=3.9×10−114x^3 = 3.9 \times 10^{-11}
  • x3=9.75×10−12x^3 = 9.75 \times 10^{-12}
  • x=2.14×10−4Mx = 2.14 \times 10^{-4} M

Molar Solubility in 0.01 M NaF

  • Ksp=[Ca2+][F−]2=(x)(0.01)2=3.9×10−11K_{sp} = [Ca^{2+}][F^-]^2 = (x)(0.01)^2 = 3.9 \times 10^{-11}

  • x=3.9×10−7Mx = 3.9 \times 10^{-7} M

  • The molar solubility decreased by a factor of 549.7.

Molar Solubility of Zinc Phosphate

  • Zn<em>3(PO</em>4)<em>2(s)⇌3Zn2+(aq)+2PO</em>43−(aq)Zn<em>3(PO</em>4)<em>2(s) \rightleftharpoons 3Zn^{2+}(aq) + 2PO</em>4^{3-}(aq), Ksp=9.1×10−33K_{sp} = 9.1 \times 10^{-33}
  • K<em>sp=[Zn2+]3[PO</em>43−]2=(3x)3(2x)2=108x5K<em>{sp} = [Zn^{2+}]^3[PO</em>4^{3-}]^2 = (3x)^3(2x)^2 = 108x^5
  • 108x5=9.1×10−33108x^5 = 9.1 \times 10^{-33}
  • x5=8.4×10−35x^5 = 8.4 \times 10^{-35}
  • x=1.53×10−7Mx = 1.53 \times 10^{-7} M

Equilibrium Constant Calculation

  • Zn2+(aq)+4CN−(aq)⇌Zn(CN)42−Zn^{2+}(aq) + 4CN^-(aq) \rightleftharpoons Zn(CN)_4^{2-}
  • Zn<em>3(PO</em>4)<em>2(s)⇌3Zn2+(aq)+2PO</em>43−(aq)Zn<em>3(PO</em>4)<em>2(s) \rightleftharpoons 3Zn^{2+}(aq) + 2PO</em>4^{3-}(aq)
  • Zn<em>3(PO</em>4)<em>2(s)+12CN−(aq)⇌3Zn(CN)</em>42−(aq)+2PO43−(aq)Zn<em>3(PO</em>4)<em>2(s) + 12CN^-(aq) \rightleftharpoons 3Zn(CN)</em>4^{2-}(aq) + 2PO_4^{3-}(aq)
  • K=K<em>f3×K</em>sp=(5.0×1011)3×9.1×10−33=1.14×101K = K<em>f^3 \times K</em>{sp} = (5.0 \times 10^{11})^3 \times 9.1 \times 10^{-33} = 1.14 \times 10^{1}

Molar Solubility of Nickel Cyanide

  • Ni(CN)<em>2(s)⇌Ni2+(aq)+2CN−(aq)Ni(CN)<em>2(s) \rightleftharpoons Ni^{2+}(aq) + 2CN^-(aq), K</em>sp=3×10−11K</em>{sp} = 3 \times 10^{-11}
  • Ni2+(aq)+6NH<em>3(aq)⇌Ni(NH</em>3)<em>62+(aq)Ni^{2+}(aq) + 6NH<em>3(aq) \rightleftharpoons Ni(NH</em>3)<em>6^{2+}(aq), K</em>f=2×108K</em>f = 2 \times 10^8
  • Ni(CN)<em>2(s)+6NH</em>3(aq)⇌Ni(NH<em>3)</em>62+(aq)+2CN−(aq)Ni(CN)<em>2(s) + 6NH</em>3(aq) \rightleftharpoons Ni(NH<em>3)</em>6^{2+}(aq) + 2CN^-(aq)
  • K=K<em>sp×K</em>f=(3×10−11)(2×108)=6×10−3K = K<em>{sp} \times K</em>f = (3 \times 10^{-11})(2 \times 10^8) = 6 \times 10^{-3}
  • K=[Ni(NH<em>3)</em>62+][CN−]2[NH3]6=(x)(2x)2(1.00)6K = \frac{[Ni(NH<em>3)</em>6^{2+}][CN^-]^2}{[NH_3]^6} = \frac{(x)(2x)^2}{(1.00)^6}
  • 6×10−3=4x36 \times 10^{-3} = 4x^3
  • x=0.114Mx = 0.114 M

Isomerization of Ethanal to Vinyl Alcohol

  • CH<em>3C(O)H⇌H</em>2CCHOHCH<em>3C(O)H \rightleftharpoons H</em>2CCHOH
  • ΔG∘=−127.6kJmol\Delta G^\circ = -127.6 \frac{kJ}{mol} for ethanal, ΔG∘=−83.02kJmol\Delta G^\circ = -83.02 \frac{kJ}{mol} for vinyl alcohol.
  • ΔGrxn∘=(1mol×−83.02kJmol)−(1mol×−127.6kJmol)=44.58kJmol\Delta G_{rxn}^\circ = (1 mol \times -83.02 \frac{kJ}{mol}) - (1 mol \times -127.6 \frac{kJ}{mol}) = 44.58 \frac{kJ}{mol}

Equilibrium Constant Calculation

  • ΔG=−RTlnK\Delta G = -RTlnK
  • lnK=−ΔGRTlnK = - \frac{\Delta G}{RT}
  • lnK=−44.58kJmol×1000J1kJ8.314Jmol⋅K×298K=−18.01lnK = - \frac{44.58 \frac{kJ}{mol} \times 1000 \frac{J}{1 kJ}}{8.314 \frac{J}{mol \cdot K} \times 298 K} = -18.01
  • K=e−18.01=1.48×10−8K = e^{-18.01} = 1.48 \times 10^{-8}