Chemistry Labs

Problem 2

Kinetic isotope effect (KIE) and zero-point vibrational energy (ZPE). The harmonic oscillator model gives the vibrational frequency ν=12πkμ\nu = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{\mu}}, where kk is the force constant and μ=m1m2m1+m2\mu = \dfrac{m_1 m_2}{m_1 + m_2} is the reduced mass. Vibrational energies are En=(n+12)hνE_n = (n + \tfrac{1}{2})h\nu (n=0,1,2,…n = 0, 1, 2, \dots), and the zero-point energy is ZPE=12hν\text{ZPE} = \tfrac{1}{2}h\nu. (a) Calculate the reduced masses μCH\mu_{\ce{CH}} and μCD\mu_{\ce{CD}} in atomic mass units (u), taking m(C)=12.00m(\ce{C}) = 12.00 u, m(H)=1.008m(\ce{H}) = 1.008 u, and m(D)=2.014m(\ce{D}) = 2.014 u. (b) Given kCH=kCDk_{\ce{CH}} = k_{\ce{CD}} and the C−H\ce{C-H} stretching wavenumber ν~CH=2900 cm−1\tilde{\nu}_{\ce{CH}} = 2900\ \text{cm}^{-1}, calculate the C−D\ce{C-D} stretching wavenumber ν~CD\tilde{\nu}_{\ce{CD}} (cm−1^{-1}). (c) Calculate ZPECH\text{ZPE}_{\ce{CH}} and ZPECD\text{ZPE}_{\ce{CD}} in kJ mol−1^{-1}. (d) Calculate the difference in bond dissociation energies ΔBDE=BDECD−BDECH\Delta \text{BDE} = \text{BDE}_{\ce{CD}} - \text{BDE}_{\ce{CH}} (kJ mol−1^{-1}). (e) Assuming Ea≈BDEE_a \approx \text{BDE} and identical Arrhenius pre-exponential factors, calculate the theoretical primary KIE kCH/kCDk_{\ce{CH}}/k_{\ce{CD}} at 25 ∘C25\ ^{\circ}\text{C}. (f) In the chromic acid oxidation of diphenylmethanol, measured first-order rate constants are kCH=0.012 min−1k_{\ce{CH}} = 0.012\ \text{min}^{-1} and kCD=0.0018 min−1k_{\ce{CD}} = 0.0018\ \text{min}^{-1}. Compare this experimental ratio with the theoretical value and determine whether C−H\ce{C-H} bond cleavage is rate-determining.
Step 3 of 6: Zero-point vibrational energies
ZPE=12hcν~NA⇒ZPECH=17.35 kJ mol−1,ZPECD=12.73 kJ mol−1\text{ZPE} = \tfrac{1}{2}hc\tilde{\nu}N_A \Rightarrow \text{ZPE}_{\ce{CH}} = 17.35\ \text{kJ mol}^{-1},\quad \text{ZPE}_{\ce{CD}} = 12.73\ \text{kJ mol}^{-1}
Analysis

Multiplying by hcNA=(6.626×10−34×2.998×1010×6.022×1023×10−3)=0.01196hcN_A = (6.626\times 10^{-34}\times 2.998\times 10^{10}\times 6.022\times 10^{23}\times 10^{-3}) = 0.01196 kJ mol−1^{-1} cm gives ZPECH=12×0.01196×2900=17.35\text{ZPE}_{\ce{CH}} = \frac{1}{2}\times 0.01196\times 2900 = 17.35 kJ mol−1^{-1} and ZPECD=12.73\text{ZPE}_{\ce{CD}} = 12.73 kJ mol−1^{-1}.