Chemistry of Life Basics

Question 1 (20)

(a) (4)

• $^{14}\text{C}$: protons $=6$, electrons $=6$, neutrons $=14-6=8$. (1)
• $^{12}\text{C}$: protons $=6$, electrons $=6$, neutrons $=6$. (1)
• Chemically identical: chemistry is determined by electron configuration, which depends on proton (electron) count = 6 in both → same valence, same bonding. (1)
• Physically distinguishable: differing neutron count → different mass and, for $^{14}\text{C}$, nuclear instability (radioactivity). (1)

(b) (4) $\bar{A} = (0.9893)(12.000) + (0.0107)(13.003)$

• $0.9893 \times 12.000 = 11.8716$ (1)
• $0.0107 \times 13.003 = 0.139132$ (1)
• Sum $= 12.01073$ (1)
• $\approx 12.01$ u (4 s.f.) (1)

(c) (6) (i) At $t=t_{1/2}$, $N=N_0/2$: $\tfrac12 = e^{-\lambda t_{1/2}} \Rightarrow \lambda = \dfrac{\ln 2}{t_{1/2}} = \dfrac{0.6931}{5730} = 1.2097\times10^{-4}\ \text{yr}^{-1}$. (3) (ii) $0.220 = e^{-\lambda t} \Rightarrow t = \dfrac{-\ln 0.220}{\lambda} = \dfrac{1.5141}{1.2097\times10^{-4}} = 12516 \approx 12520$ years. (3)

(d) (6)

import math
def carbon_age(fraction, half_life=5730):
    lam = math.log(2) / half_life      # decay constant from half-life
    return -math.log(fraction) / lam   # invert N/N0 = e^(-lam t)

• Line 1: derive $\lambda$ (1). Line 2: solve for $t$ (1). Correct math structure (1).
• Carbon centrality: 4 valence electrons → forms 4 stable covalent bonds, enabling long chains, rings, branches → vast diversity of biomolecules; isotope identity does not change valence, so all carbon isotopes bond identically. (3)

Question 2 (22)

(a) (6)

• Ionic: electron transfer, electrostatic attraction of ions; strong (~150–400 kJ/mol). (1)
• Covalent: electron sharing; strongest (~150–1000 kJ/mol). (1)
• Hydrogen bond: weak dipole attraction between H (bonded to N/O/F) and lone pair; weak (~4–40 kJ/mol). (1)
• O–H bond in water: polar covalent — $\Delta\text{EN} = 3.44-2.20 = 1.24$ → shared but unequal, partial charges. (1.5)
• O···H between molecules: hydrogen bond — attraction of H($\delta^+$) to O($\delta^-$) lone pair on neighbour; not electron sharing. (1.5)

(b) (8) Water: $\Delta T = \dfrac{Q}{mc} = \dfrac{8.37\times10^{8}}{(5.00\times10^{3})(4186)} = \dfrac{8.37\times10^{8}}{2.093\times10^{7}} = 39.99 \approx 40.0\ \text{K}$. (3) Iron: $\Delta T = \dfrac{8.37\times10^{8}}{(5.00\times10^{3})(449)} = \dfrac{8.37\times10^{8}}{2.245\times10^{6}} = 372.8\ \text{K}$. (3) Consequence: water's high specific heat means it warms/cools ~9× more slowly than iron → stabilises organism and environmental temperatures, buffers cells against thermal shock, moderates climate. (2)

(c) (8) $h = \frac{2(0.0728)(\cos 0^\circ)}{(1000)(9.81)(2.00\times10^{-5})}$

• Numerator $= 2\times0.0728 = 0.1456$ (1)
• Denominator $= 1000\times9.81\times2.00\times10^{-5} = 0.1962$ (2)
• $h = 0.1456/0.1962 = 0.742\ \text{m}$ (2)
• Interpretation: adhesion (water–tube attraction) pulls water up; cohesion (water–water H-bonds) maintains a continuous column; surface tension at the meniscus sustains the pull → together drive capillary rise in xylem, aiding water transport up plants. (3)

Question 3 (18)

(a) (4)

• $\text{pH} = -\log_{10}[\text{H}^+]$; scale $0$–$14$. (2)
• $\text{pH} = -\log_{10}(4.0\times10^{-8}) = 8 - \log_{10}4.0 = 8 - 0.602 = 7.40$. (1)
• Since $[\text{H}^+]=4.0\times10^{-8} < 10^{-7}$ → basic (pH $7.40 > 7$). (1)

(b) (4) $\text{pH} = 6.10 + \log_{10}\frac{24}{1.2} = 6.10 + \log_{10}20 = 6.10 + 1.301 = 7.40$ (4)

(c) (6)

• $\text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3$ (reactants CO₂+water; product carbonic acid). (1.5)
• $\text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^-$. (1.5)
• Adding strong acid raises $[\text{H}^+]$; by Le Chatelier the equilibrium shifts left, $\text{HCO}_3^-$ consumes excess $\text{H}^+$ to form $\text{H}_2\text{CO}_3$ (then CO₂ exhaled), so pH change is minimised. (3)

(d) (4)

import math
def blood_pH(HCO3, H2CO3, pKa=6.10):
    return pKa + math.log10(HCO3 / H2CO3)
 
def is_homeostatic(pH):
    return 7.35 <= pH <= 7.45

• Correct HH implementation (1.5), correct range check (1.5).
• Enzymes have optimal pH; deviations alter ionisation of active-site groups → denaturation/loss of activity, so buffers preserve function. (1)

[
  {"claim":"Average atomic mass of carbon rounds to 12.01 u", "code":"m = 0.9893*12.000 + 0.0107*13.003; result = abs(m - 12.01) < 0.005"},
  {"claim":"C-14 fossil at 22% is about 12520 years old", "code":"lam = log(2)/5730; t = -log(0.220)/lam; result = abs(float(t) - 12520) < 15"},
  {"claim":"Pond delta T is 40.0 K", "code":"dT = 8.37e8/(5.00e3*4186); result = abs(float(dT) - 40.0) < 0.2"},
  {"claim":"Capillary rise is about 0.742 m", "code":"h = (2*0.0728*1)/(1000*9.81*2.00e-5); result = abs(float(h) - 0.742) < 0.01"},
  {"claim":"Blood pH via Henderson-Hasselbalch is 7.40", "code":"pH = 6.10 + log(Rational(24,12)*10,10).evalf() if False else 6.10 + float(log(20,10)); result = abs(pH - 7.40) < 0.01"}
]