1.(5 pts) The equilibrium constant Kc for the reaction: Ha(g) + Br2(g) «* 2HBr(g) is 2.18 x 10% at 730°C. Starting with 3.20 moles of HBr in a 12.0-L reaction vessel, calculate the concentrations of H2, Br2, and HBr at equilibrium. 2. (10 pts) The following equilibrium process has been studied at 230°C: 2N0(g) + 02(g) + 2N02(g) In one experiment, the concentrations of the reacting species at equilibrium are found to be [NO] = 0.0542 M, [02] = 0.127 M, and [NO2] = 15.5 M. Calculate the equilibriu
Answer (1)
Problem 1. H₂ + Br₂ ⇌ 2HBr EquilibriumData ProvidedKc = 2.18 × 10⁶ at 730°CStarting amount: 3.20 moles of HBrVolume: 12.0 LNeed to find: [H₂], [Br₂], and [HBr] at equilibriumSince we're starting with HBr and it can decompose backwards, we need to consider the reverse reaction:2HBr(g) ⇌ H₂(g) + Br₂(g)Initial concentration of HBr = 3.20 mol ÷ 12.0 L = 0.267 M[tex]\left[\begin{array}{cccccc}&2HBr&forward/back&H_{2}&+&Br_{2} \\I (M)&0.267&forward&0&&0\\C(M)&-2x&forward&0&&0\\E(M)&0.267-2x&&x&&x\end{array}\right][/tex]Write the equilibrium expression.For the forward reaction H₂ + Br₂ ⇌ 2HBr:Kc = [HBr]²/([H₂][Br₂])For the reverse reaction 2HBr ⇌ H₂ + Br₂:K'c = 1/Kc = 1/(2.18 × 10⁶) = 4.59 × 10⁻⁷Write the equilibrium expressionFor the forward reaction H₂ + Br₂ ⇌ 2HBr:Kc = [HBr]²/([H₂][Br₂])For the reverse reaction 2HBr ⇌ H₂ + Br₂:K'c = 1/Kc = 1/(2.18 × 10⁶) = 4.59 × 10⁻⁷Set up the equilibrium expression with variablesK'c = [H₂][Br₂]/[HBr]²4.59 × 10⁻⁷ = (x)(x)/(0.267-2x)²4.59 × 10⁻⁷ = x²/(0.267-2x)²Solve for xSince Kc is very large (2.18 × 10⁶), K'c is very small (4.59 × 10⁻⁷), meaning very little HBr will decompose. We can assume 2x << 0.267.4.59 × 10⁻⁷ ≈ x²/(0.267)²4.59 × 10⁻⁷ = x²/0.0713x² = 4.59 × 10⁻⁷ × 0.0713x² = 3.27 × 10⁻⁸x = 1.81 × 10⁻⁴ MCalculate equilibrium concentrations.[H₂] = x = 1.81 × 10⁻⁴ M[Br₂] = x = 1.81 × 10⁻⁴ M[HBr] = 0.267 - 2x = 0.267 - 2(1.81 × 10⁻⁴) = 0.267 - 0.000362 = 0.267 MVerify the assumption.2x = 2(1.81 × 10⁻⁴) = 3.62 × 10⁻⁴This is much smaller than 0.267, so our assumption was valid.Final Answers[H₂] = 1.81 × 10⁻⁴ M[Br₂] = 1.81 × 10⁻⁴ M[HBr] = 0.267 MProblem 2: 2NO + O₂ ⇌ 2NO₂ EquilibriumData ProvidedTemperature: 230°C[NO] = 0.0542 M[O₂] = 0.127 M[NO₂] = 15.5 MNeed to find: KcWrite the balanced equation2NO(g) + O₂(g) ⇌ 2NO₂(g)Write the equilibrium constant expressionFor the reaction aA + bB ⇌ cC + dD:Kc = [C]ᶜ[D]ᵈ/[A]ᵃ[B]ᵇFor our reaction:Kc = [NO₂]²/([NO]²[O₂])Substitute the equilibrium concentrationsKc = (15.5)²/((0.0542)²(0.127))Calculate step by stepNumerator: [NO₂]² = (15.5)² = 240.25Denominator: [NO]² = (0.0542)² = 0.002937Denominator: [NO]²[O₂] = 0.002937 × 0.127 = 0.000373Final calculationKc = 240.25/0.000373 = 644,000Final Answer is Kc = 6.44 × 10⁵
