Unit 3: Energy & Rates of Reaction

Assessment AS Learning — Practice Quiz

Not graded

Purpose: Self-check thermochemistry (calorimetry, Hess) and kinetics (rate laws, activation energy).

Calorimetry

25.0 g of water is heated from 20.0 °C to 80.0 °C. How much heat absorbed? (c = 4.18 J/g·°C)

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q = mcΔT = (25.0)(4.18)(60.0) = 6270 J = 6.27 kJ.

A reaction releases 12.5 kJ heating 250 g of water in a calorimeter. What's ΔT? (c = 4.18 J/g·°C)

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ΔT = q/(mc) = 12500 / (250 × 4.18) = 11.96 °C.

Hess's Law & ΔH_f

Using Hess's law: A → B (ΔH = −50 kJ); B → C (ΔH = +30 kJ). Find ΔH for A → C.

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ΔH = (−50) + (+30) = −20 kJ.

For CH₄(g) + 2 O₂ → CO₂ + 2 H₂O(g), use ΔH_f°: CH₄ = −74.8, CO₂ = −393.5, H₂O(g) = −241.8 kJ/mol. Find ΔH°_rxn.

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ΔH = [(−393.5) + 2(−241.8)] − [(−74.8) + 0] = (−877.1) − (−74.8) = −802.3 kJ/mol.

Reaction Rates

Given rate data: when [A] doubles, rate quadruples; when [B] doubles, rate doesn't change. Order in A and B?

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A is second order (rate ∝ [A]²); B is zero order. Rate law: rate = k[A]².

For rate = k[A]²[B], units of k?

M⁻¹ s⁻¹
M⁻² s⁻¹
s⁻¹
M s⁻¹

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For a third-order rate law: k = M/s ÷ M³ = M⁻² s⁻¹.

A rate doubles when temperature rises 10 °C. Identify two reasons (collision theory).

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More molecules have KE > E_a; collision frequency increases.

Activation Energy

Sketch a potential-energy diagram for an exothermic reaction with ΔH = −100 kJ and E_a = 50 kJ.

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Reactants high → activated complex (peak at +50 kJ above reactants) → products 100 kJ below reactants. Forward E_a = 50; reverse E_a = 150 kJ.

Using Arrhenius: at 300 K rate constant k₁ = 2.0×10⁻³ s⁻¹; at 320 K k₂ = 6.0×10⁻³ s⁻¹. Estimate E_a. R = 8.314 J/mol·K.

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$$\ln(k_2/k_1) = -\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$$ ln(3) = 1.099. (1/320−1/300) = −2.083×10⁻⁴. So E_a = (1.099 × 8.314)/2.083×10⁻⁴ ≈ 4.39×10⁴ J/mol = 43.9 kJ/mol.

Q10

A catalyst increases rate by ___?

Increasing E_a
Lowering E_a via alternate pathway
Changing ΔH

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Catalyst lowers E_a via an alternate mechanism; ΔH unchanged.

Q11

Identify the rate-determining step: Step 1 (slow): A + B → C; Step 2 (fast): C + D → E. Predict rate law.

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Slow step is rate-determining → rate = k[A][B].

Q12

For C(s) + O₂(g) → CO₂(g), ΔH = −393.5 kJ/mol. Calculate energy released when 12.0 g of carbon burns.

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n(C) = 12.0/12.01 = 1.00 mol → q = 1.00 × 393.5 = 393.5 kJ released.