1972 Ap Chemistry Free Response Answers ((link)) Today

(a) What is the rate equation for the reaction? (b) What is the numerical value of the rate constant k? What are its dimensions? (c) Propose a reaction mechanism for this reaction.

Given the following thermochemical equations at 25°C:

ΔG∘=−nFEcell∘cap delta cap G raised to the composed with power equals negative n cap F cap E sub c e l l end-sub raised to the composed with power = Standard Gibbs Free Energy change (in Joules) 1972 ap chemistry free response answers

Let ( s ) = molar solubility. ( K_sp = [Pb^2+][F^-]^2 = (s)(2s)^2 = 4s^3 ) ( 4s^3 = 3.7 \times 10^-8 ) ( s^3 = 9.25 \times 10^-9 ) ( s = \sqrt[3]9.25 \times 10^-9 ) ( s \approx 2.10 \times 10^-3 , \textM )

The Free Response section, often called the "essay" section, required students to show their work explicitly and explain their reasoning. Key Topics Covered in 1972 Free Response Questions (a) What is the rate equation for the reaction

PV=nRT=mMRTcap P cap V equals n cap R cap T equals the fraction with numerator m and denominator cap M end-fraction cap R cap T Rearranging the formula to solve for

: 1-bromo-1-chloroethane contains a chiral center (carbon bonded to CH3cap C cap H sub 3 ), allowing for enantiomers. Ethene ( ) : Geometric Isomers : cis- and trans-1-bromo-2-chloroethene. (c) Propose a reaction mechanism for this reaction

| Experiment | [A] (mole·liter⁻¹) | [B] (mole·liter⁻¹) | Initial Rate of Formation of C (mole·liter⁻¹·min⁻¹) | | :--- | :--- | :--- | :--- | | 1 | 0.60 | 0.15 | 6.3×10⁻³ | | 2 | 0.20 | 0.60 | 2.8×10⁻³ | | 3 | 0.20 | 0.15 | 7.0×10⁻⁴ |

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First, total moles of HCl initially added are 0.100 L × 2.00 M = 0.200 mol. The excess HCl that did not react is then determined via titration with NaOH. From the titration, moles of excess HCl are 0.0866 L × 1.50 M = 0.130 mol. Therefore, the moles of HCl that reacted with the basic components of the mixture is 0.200 mol - 0.130 mol = 0.070 mol. Since the K₂CO₃ from part (a) consumed 0.020 mol of HCl, the remaining 0.050 mol of HCl must have reacted with the KOH (KOH + HCl → KCl + H₂O). This means there are 0.050 mol of KOH in the original sample, which corresponds to a mass of 2.81 g. The percentage of KOH is therefore (2.81 g / 5.00 g) × 100% = 56.1% . The remaining mass is KCl: 5.00 g - (1.38 g + 2.81 g) = 0.81 g. The percentage of KCl is (0.81 g / 5.00 g) × 100% = 16.3% .

: Includes geometric (cis/trans) isomers and structural isomers. Energy & Electrochemistry One question focused on calculating free energy ( ΔGcap delta cap G ) and enthalpy ( ΔHcap delta cap H ) using electrochemistry data. Key Formula : Calculation : For a specific redox reaction yielding ΔGcap delta cap G was determined to be