Chemistry in Everyday Life (compressed)

Question 1

(a) [3]

• Antacid $\text{Mg(OH)}_2$ acts chemically after acid is secreted: the hydroxide directly neutralises already-present $\text{HCl}$ (Brønsted acid–base) [1].
• $\text{H}_2$-receptor antagonists (ranitidine) act before/upstream: they block histamine $\text{H}_2$ receptors on parietal cells, preventing acid secretion rather than neutralising it [1].
• Thus antacid = symptomatic neutralisation of existing acid; receptor antagonist = reduces the amount of acid produced [1].

(b) [4]

• Moles $\text{Mg(OH)}_2 = 0.350/58.3 = 6.004\times10^{-3}\ \text{mol}$ [1].
• Stoichiometry $1:2$, so moles $\text{HCl} = 2\times6.004\times10^{-3} = 1.201\times10^{-2}\ \text{mol}$ [1].
• Volume $= n/C = 1.201\times10^{-2}/0.100 = 0.1201\ \text{L}$ [1].
• $= \mathbf{120.1\ mL}$ (≈120 mL) [1].

(c) [4]

• $B_0 = 6.004\times10^{-3}$ mol base; each mole of diprotic base consumes 2 mol acid, so acid capacity $= 2B_0$ mol [1].
• Acid supplied by time $t$: $n_{\text{acid}}=r\,t$ [1].
• Base fully consumed when $r\,t^* = 2B_0 \Rightarrow t^* = \dfrac{2B_0}{r}$ [1].
• $t^* = \dfrac{2(6.004\times10^{-3})}{2.0\times10^{-3}} = 6.004\ \text{min} \approx \mathbf{6.0\ min}$. ($V$ is a distractor — not needed) [1].

(d) [4]

• For diprotic base of mass $m$: moles $=m/M_d$, acid capacity $=2m/M_d$, so $t^*_{\text{di}} = \dfrac{2m}{M_d\,r}$ [1].
• For monoprotic base of equal mass: moles $=m/M_m$, capacity $=m/M_m$, so $t^*_{\text{mono}} = \dfrac{m}{M_m\,r}$ [1].
• Ratio $\dfrac{t^*_{\text{di}}}{t^*_{\text{mono}}} = \dfrac{2m/(M_d r)}{m/(M_m r)} = \dfrac{2M_m}{M_d}$ (QED) [1].
• Drawback of $\text{NaHCO}_3$: releases $\text{CO}_2$ causing belching/distension, and being alkaline can cause systemic alkalosis / rebound acid secretion [1].

Question 2

(a) [4]

• Soap: anionic hydrophilic head ($\text{–COO}^-$) [1].
• Anionic detergent: negatively charged head, e.g. sodium lauryl (dodecyl) sulphate, $\text{CH}_3(\text{CH}_2)_{11}\text{OSO}_3^-\text{Na}^+$ [1].
• Cationic detergent: positively charged head (quaternary ammonium), e.g. cetyltrimethylammonium bromide [1].
• Non-ionic detergent: no charge, e.g. polyethylene-glycol ester of a fatty acid / Tween/stearate–PEG [1].

(b) [4]

• Moles tristearin $= 100/891 = 0.11223\ \text{mol}$ [1].
• Each mole gives 3 mol sodium stearate: $0.11223\times3 = 0.33670\ \text{mol}$ [1].
• Mass $= 0.33670\times306 = 103.03\ \text{g}$ [1].
• $\approx \mathbf{103\ g}$ of soap [1].

(c) [3]

• $\Delta G_{\text{mic}} = RT\ln(\text{CMC}) = 8.314\times298\times\ln(10^{-3})$ [1].
• $\ln(10^{-3}) = -6.9078$; $\Delta G = 8.314\times298\times(-6.9078) = -17114\ \text{J mol}^{-1}$ [1].
• $\approx \mathbf{-17.1\ kJ\,mol^{-1}}$; negative sign ⇒ micelle formation is spontaneous (favourable) at this concentration [1].

(d) [4]

• Let $\text{CMC}_2 = \text{CMC}_1/10$. Then $\Delta G_2 - \Delta G_1 = RT[\ln(\text{CMC}_1/10) - \ln(\text{CMC}_1)] = RT\ln(1/10) = -RT\ln 10.$ [2]
• Numerically $-RT\ln10 = -8.314\times298\times2.3026 = -5705\ \text{J mol}^{-1} \approx \mathbf{-5.7\ kJ\,mol^{-1}}$ [1].
• Longer hydrocarbon tails increase hydrophobicity ⇒ stronger drive to leave water ⇒ lower CMC ⇒ more negative $\Delta G_{\text{mic}}$; each $-CH_2-$ adds a roughly constant favourable increment [1].

Question 3

(a) [3]

• Oxidative rancidity is a radical chain reaction: initiation forms $R^\bullet$, propagation ($R^\bullet + O_2 \to ROO^\bullet$, etc.) multiplies radicals [1].
• A chain-breaking antioxidant donates H to quench $ROO^\bullet$, terminating the chain, so one molecule stops many propagation cycles — more effective than merely limiting $O_2$, which slows but does not break chains already started [1].
• Water-soluble food antioxidant: ascorbic acid (vitamin C) [1].

(b) [5]

• Separate variables: $\dfrac{d[P]}{[P](P_{\max}-[P])} = k\,dt$ [1].
• Partial fractions: $\dfrac{1}{[P](P_{\max}-[P])} = \dfrac{1}{P_{\max}}\left(\dfrac{1}{[P]} + \dfrac{1}{P_{\max}-[P]}\right)$ [1].
• Integrate: $\dfrac{1}{P_{\max}}\ln\dfrac{[P]}{P_{\max}-[P]} = k t + C$ [1].
• Apply $[P](0)=P_0$: $C = \dfrac{1}{P_{\max}}\ln\dfrac{P_0}{P_{\max}-P_0}$ [1].
• Solve for $[P]$: $[P](t) = \dfrac{P_{\max}}{1 + \left(\frac{P_{\max}-P_0}{P_0}\right)e^{-kP_{\max}t}}$ (QED) [1].

(c) [4] Closed form (invert the solution for $t$ at $[P]=T$): $t = \frac{1}{kP_{\max}}\ln\!\left[\frac{T}{P_{\max}-T}\cdot\frac{P_{\max}-P_0}{P_0}\right].$ [2]

import math
def shelf_life(k, Pmax, P0, threshold):
    T = threshold
    return (1.0/(k*Pmax)) * math.log((T/(Pmax - T)) * ((Pmax - P0)/P0))

[2] (correct implementation of the analytic formula)

(d) [3] Using $P_{\max}=1$, $P_0=0.01$, $T=0.5$:

• Argument $= \dfrac{0.5}{0.5}\cdot\dfrac{0.99}{0.01} = 99$; $\ln 99 = 4.5951$ [1].
• Untreated: $t = \dfrac{4.5951}{0.20\times1} = 22.98\ \text{days} \approx 23\ \text{days}$ [1].
• Treated: $t = \dfrac{4.5951}{0.05\times1} = 91.90\ \text{days} \approx 92\ \text{days}$; extension factor $=0.20/0.05 = \mathbf{4\times}$ [1].

[
  {"claim":"Q1b: HCl volume ~120.1 mL","code":"n=0.350/58.3; nHCl=2*n; V_L=nHCl/0.100; result = abs(V_L*1000-120.1)<0.5"},
  {"claim":"Q1c: neutralisation time ~6.0 min","code":"B0=0.350/58.3; r=2.0e-3; t=2*B0/r; result = abs(t-6.0)<0.1"},
  {"claim":"Q2b: soap mass ~103 g","code":"n=100/891; mass=3*n*306; result = abs(mass-103.0)<1.0"},
  {"claim":"Q2c: dGmic ~ -17.1 kJ/mol","code":"import sympy as sp; dG=8.314*298*sp.log(1e-3); result = abs(float(dG)/1000 - (-17.1))<0.2"},
  {"claim":"Q2d: increment ~ -5.7 kJ/mol","code":"import sympy as sp; inc=-8.314*298*sp.log(10); result = abs(float(inc)/1000 - (-5.7))<0.1"},
  {"claim":"Q3d: shelf life ratio is 4 and untreated ~23 days","code":"import sympy as sp; arg=(0.5/0.5)*(0.99/0.01); t1=float(sp.log(arg))/(0.20*1); t2=float(sp.log(arg))/(0.05*1); result = abs(t1-22.98)<0.2 and abs(t2/t1-4)<1e-6"}
]