Thermodynamics
A reaction has ΔH = −85 kJ/mol and ΔS = +120 J/(mol·K) at 298 K. Calculate ΔG and determine whether the reaction is spontaneous.
G = ΔH − TΔS = −85,000 − (298)(120) = −85,000 − 35,760 = −120,760 J/mol ≈ −120.8 kJ/mol. Negative ΔG → spontaneous at 298 K.
A weak acid HA has Ka = 1.8 × 10⁻⁵. Calculate the pH of a 0.10 M solution of HA.
Ka = x²/0.10 → x² = 1.8×10⁻⁶ → x = 1.34×10⁻³ M = [H⁺]. pH = −log(1.34×10⁻³) ≈ 2.87
For the reaction 2NO(g) + O₂(g) → 2NO₂(g), the rate of disappearance of NO is 5.0×10⁻⁵ Ms⁻¹. What is the rate of disappearance of O₂?
2.5×10⁻⁵ Ms⁻¹ Because two molecules of NO are consumed per one molecule of O₂, the rate of O₂ disappearance is half that of NO.
For a reaction where ΔH < 0 and ΔS < 0, at what temperature range is the reaction spontaneous?
Spontaneous only at LOW temperatures, where the −TΔS term is small and doesn't overcome the negative ΔH. At high T, −TΔS becomes large and positive, making ΔG positive.
For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Kc = 0.50 at 400°C. If Qc = 2.0, predict which direction the reaction will shift and explain why.
Qc (2.0) > Kc (0.50), so the reaction is right of equilibrium. It will shift LEFT (toward reactants) to decrease Q and reach equilibrium.
For the reaction 2NO(g) + Cl₂(g) → 2NOCl(g), experiments show the rate quadruples when [NO] is doubled, and the rate doubles when [Cl₂] is doubled. What is the rate law?
[NO]²[Cl₂] — Second order in NO, first order in Cl₂, third order overall.
Explain why dissolving NH₄NO₃ in water feels cold, using thermodynamic reasoning.
Dissolving NH₄NO₃ is endothermic (ΔH > 0); it absorbs heat from the surroundings. ΔS is positive (solid → ions in solution), making ΔG negative at room temperature (entropy-driven).
Explain the buffer capacity of a CH₃COOH/CH₃COO⁻ buffer and calculate the pH after adding 0.01 mol HCl to 1 L of a buffer with 0.20 M each of acetic acid and acetate. Ka = 1.8×10⁻⁵
pH = pKa + log([A⁻]/[HA]). After adding 0.01 mol HCl: [HA]=0.21, [A⁻]=0.19. pH = 4.74 + log(0.19/0.21) = 4.74 − 0.044 = 4.70. Buffer resisted large pH drop.
The conversion of cyclopropane (0.25 M) to propene is first order with k = 6.7×10⁻⁴ s⁻¹ at 500°C. How long will it take to convert 74% of the starting material?
2011 seconds Using ln[Aₜ/A₀] = −kt → ln(0.26) = −6.7×10⁻⁴ × t → t = 1.35/6.7×10⁻⁴ ≈ 2011 s.
The standard enthalpy of formation of liquid water is −285.8 kJ/mol. Using Hess's Law, calculate ΔH for: 2H₂O(g) → 2H₂(g) + O₂(g), given ΔH°vap(H₂O) = 44 kJ/mol.
H₂(g) + ½O₂(g) → H₂O(l) ΔH = −285.8 kJ. H₂O(l) → H₂O(g) ΔH = +44 kJ/mol. So H₂(g) + ½O₂(g) → H₂O(g) ΔH = −241.8 kJ. Reverse×2: 2H₂O(g) → 2H₂ + O₂ ΔH = +483.6 kJ
Which of the following pairs of mathematical expressions can be used to correctly calculate the pH and pOH of a 0.0015 M KOH(aq) solution at 25°C?
pOH = −log(0.0015)
pH = 14 − pOH
The reaction between NO₂ and CO produces NO and CO₂ via two steps: Step 1: NO₂ + NO₂ → NO + NO₃ (slow); Step 2: NO₃ + CO → NO₂ + CO₂ (fast). Write the overall equation, identify the intermediate, and state the rate law.
Overall: NO₂ + CO → NO + CO₂. Intermediate: NO₃ (produced in step 1, consumed in step 2). Rate law: rate = k[NO₂]²
A calorimeter contains 250 g of water at 22.0°C. A metal sample releases 3,500 J of heat into the water. What is the final temperature? (cwater = 4.18 J/g·°C)
q = mcΔT → 3500 = (250)(4.18)(ΔT) → ΔT = 3500/1045 = 3.35°C. Final T = 22.0 + 3.35 = 25.35°C ≈ 25.4°C
The equilibrium reaction shown above represents the partial ionization of the weak acid HCN(aq). A 0.200 M HCN(aq) solution has a pH≈4.95. If 0.05g (0.010mol) of NaCN(s) is added to 100mL of 0.200 M HCN(aq), which of the following explains how and why the pH of the solution changes?
The pH will be higher than 4.95 because adding CN⁻ shifts equilibrium left via the common ion effect, reducing [H₃O⁺] and raising the pH.
For the reaction Br₂(g) + 2NO(g) → 2NOBr(g), the proposed mechanism is: Step 1: NO + Br₂ ⇌ NOBr₂ (fast equilibrium); Step 2: NOBr₂ + NO → 2NOBr (slow). What is the rate law consistent with this mechanism?
rate = k[Br₂][NO]² The slow step gives rate = k[NOBr₂][NO]. Using the pre-equilibrium from step 1: [NOBr₂] = K[NO][Br₂], substituting gives rate = k[Br₂][NO]².