BE Civil Engineering (IOE, TU) Engineering Chemistry (IOE, SH 403) Question Paper 2077 Nepal

Nernst equation. For an electrode/cell at temperature $T$:

$$E = E^{\circ} - \frac{RT}{nF}\ln Q = E^{\circ} - \frac{0.0591}{n}\log_{10} Q \quad (\text{at } 298\ \text{K})$$

• $E$ = actual (non-standard) potential, $E^{\circ}$ = standard potential
• $R$ = gas constant (8.314 J K$^{-1}$ mol$^{-1}$), $T$ = temperature (K), $F$ = Faraday constant (96485 C mol$^{-1}$)
• $n$ = number of electrons transferred, $Q$ = reaction quotient

(a) Cell reaction. Zn has the more negative $E^{\circ}$, so it is oxidised (anode); Cu$^{2+}$ is reduced (cathode).

• Anode (oxidation): $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$
• Cathode (reduction): $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$
• Overall: $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$, with $n = 2$.

(b) Standard cell EMF.

$$E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = (+0.34) - (-0.76) = \mathbf{1.10\ V}$$

(c) Actual EMF (Nernst). The reaction quotient is

$$Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{0.01}{0.10} = 0.10$$ $$E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{2}\log_{10}(0.10) = 1.10 - \frac{0.0591}{2}(-1) = 1.10 + 0.02955$$ $$\boxed{E_{\text{cell}} = \mathbf{1.1296\ V} \approx 1.13\ V}$$

(d) Feasibility. Since $E_{\text{cell}} > 0$, $\Delta G = -nFE_{\text{cell}} < 0$, so the cell reaction is spontaneous (feasible) in the written direction.

(a) Corrosion and electrochemical theory.

Corrosion is the gradual destruction/deterioration of a metal by chemical or electrochemical reaction with its environment (e.g. rusting of iron). In the wet/electrochemical theory, a metal surface in contact with an electrolyte sets up tiny galvanic cells with anodic and cathodic zones.

For iron in neutral, aerated water:

• Anode (oxidation): $\text{Fe} \rightarrow \text{Fe}^{2+} + 2e^-$
• Cathode (oxygen-absorption type): $\tfrac{1}{2}\text{O}_2 + \text{H}_2\text{O} + 2e^- \rightarrow 2\text{OH}^-$
• Combination: $\text{Fe}^{2+} + 2\text{OH}^- \rightarrow \text{Fe(OH)}_2$

Ferrous hydroxide is further oxidised by dissolved O$_2$:

$$4\text{Fe(OH)}_2 + \text{O}_2 + 2\text{H}_2\text{O} \rightarrow 4\text{Fe(OH)}_3 \xrightarrow{\;-\,\text{H}_2\text{O}\;} \text{Fe}_2\text{O}_3\cdot x\text{H}_2\text{O (rust)}$$

  O2 + H2O          electrolyte (water film)
  -------(cathode)------(anode)-------
     OH-   <----e----   Fe -> Fe2+
  ===================== IRON =========

(b) Cathodic protection — two methods.

(c) Small anode / large cathode. Corrosion current is roughly fixed by the cathodic reaction over the large cathode area. When this current is concentrated at a small anode, the current density (corrosion rate per unit area) at the anode is very high, so the small anodic region is rapidly eaten away (deep pitting/perforation). Example: a steel rivet (small) in a large copper plate — the steel rivet corrodes severely. Conversely, a copper rivet in a large steel plate is comparatively safe.

(a) Why CaCO$_3$ equivalent. Different hardness-causing salts have different molar masses, so their amounts are not directly comparable. CaCO$_3$ is chosen as a common reference because its molar mass is a convenient 100 g/mol and it is the principal scale-forming compound. Expressing every salt as an equivalent amount of CaCO$_3$ lets all contributions be added on one scale.

$$\text{CaCO}_3\text{ equiv (mg/L)} = \text{amount (mg/L)} \times \frac{100}{\text{molar mass of salt}}$$

(b) Conversion to CaCO$_3$ equivalent.

(c) Hardness.

• Temporary hardness (bicarbonates) $= 25.0 + 25.0 = \mathbf{50.0\ mg/L\ as\ CaCO_3}$
• Permanent hardness (CaSO$_4$ + MgCl$_2$) $= 20.0 + 20.0 = \mathbf{40.0\ mg/L\ as\ CaCO_3}$
• Total hardness $= 50.0 + 40.0 = \mathbf{90.0\ mg/L\ as\ CaCO_3}$

Note: NaCl is excluded because sodium salts are soluble and do not contribute to hardness.

(a) Definitions. A polymer is a high-molar-mass macromolecule formed by the repeated linking of many small repeating units (monomers). The degree of polymerization (DP) is the number of monomer (repeat) units in one polymer chain:

$$\text{DP} = \frac{\text{molar mass of polymer}}{\text{molar mass of monomer (repeat unit)}}$$

(b) Addition vs condensation polymerization.

(c) Degree of polymerization.

$$\text{DP} = \frac{2.80 \times 10^{5}}{104} = 2692.3$$ $$\boxed{\text{DP} \approx \mathbf{2692\ repeat\ units}}$$

(d) Thermoplastics vs thermosetting plastics.

(a) Gross vs net calorific value.

• Gross (Higher) Calorific Value, HCV: total heat liberated when unit mass of fuel is completely burnt and the products (including water vapour from combustion of H$_2$) are cooled back to room temperature, so the latent heat of condensation of water is included.
• Net (Lower) Calorific Value, LCV: heat available when the combustion water remains as vapour (latent heat not recovered). Hence $\text{LCV} < \text{HCV}$.

(b) Gross calorific value (bomb calorimeter).

Heat absorbed $=$ heat released by fuel. With specific heat of water $= 1$ cal g$^{-1}$ °C$^{-1}$:

$$\text{HCV} = \frac{(W + w)\,\Delta T}{m}$$

where $W$ = mass of water = 1900 g, $w$ = water equivalent = 600 g, $\Delta T = 2.4$ °C, $m = 0.75$ g.

$$\text{HCV} = \frac{(1900 + 600)\times 2.4}{0.75} = \frac{2500 \times 2.4}{0.75} = \frac{6000}{0.75} = 8000\ \text{cal/g}$$

Since $1$ cal/g $=$ $1$ kcal/kg:

$$\boxed{\text{HCV} = \mathbf{8000\ kcal/kg}}$$

(c) Net calorific value.

Mass of water produced per kg fuel $= 9 \times (\text{fraction of H}) = 9 \times 0.06 = 0.54$ kg.

$$\text{Heat lost as latent heat} = 0.54 \times 587 = 316.98\ \text{kcal/kg}$$ $$\text{LCV} = \text{HCV} - 0.09 \times \%\text{H} \times 587 = 8000 - 0.09\times 6 \times 587$$ $$\text{LCV} = 8000 - 316.98 = \boxed{\mathbf{7683.02\ kcal/kg}}$$

Catalysis is the phenomenon in which the rate of a chemical reaction is altered (usually increased) by a substance — the catalyst — that itself remains chemically unchanged in mass and composition at the end of the reaction.

Homogeneous vs heterogeneous catalysis.

Characteristics of a catalyst.

1. Small amount is sufficient; it is recovered unchanged.
2. It does not change the position of equilibrium (only speeds attainment); it lowers the activation energy.
3. It is generally specific in action and may need a promoter; its activity can be destroyed by a poison.

Haber process catalyst: finely divided iron (Fe) with molybdenum or K$_2$O/Al$_2$O$_3$ as promoter, for $N_2 + 3H_2 \rightleftharpoons 2NH_3$.

Raw materials: limestone/chalk (source of CaO), clay/shale (source of SiO$_2$, Al$_2$O$_3$, Fe$_2$O$_3$), and a small amount of gypsum added after clinkering.

Bogue compounds (main constituents):

Setting and hardening (hydration). When water is added, the silicates and aluminates hydrate and hydrolyse, forming a colloidal calcium silicate hydrate (C–S–H) gel plus Ca(OH)$_2$:

$$2(3\text{CaO·SiO}_2) + 6\text{H}_2\text{O} \rightarrow 3\text{CaO·2SiO}_2\cdot 3\text{H}_2\text{O} + 3\text{Ca(OH)}_2$$

The gel first stiffens (setting), then the interlocking crystalline hydrates grow and bind, giving strength over weeks/months (hardening).

Role of gypsum (CaSO$_4$·2H$_2$O): it reacts with C$_3$A to form ettringite and retards the flash setting of C$_3$A, giving a workable, controlled setting time.

Paint is a mechanical dispersion (suspension) of one or more finely divided pigments in a liquid medium (vehicle); when applied as a thin film it dries to give a decorative and protective coating on a surface.

Essential constituents and functions.

Varnish vs paint. A varnish is a homogeneous transparent/translucent solution of a resin (and drying oil) in a solvent — it contains no pigment. It is used to give a glossy protective, transparent film that shows the natural surface (e.g. wood grain), whereas paint is opaque and coloured because it contains pigment.

Explosive: a substance (or mixture) which, on receiving a suitable stimulus (heat, shock, friction, spark), undergoes a very rapid self-sustaining chemical reaction producing a large volume of hot gas and a sudden release of energy.

Classification with examples.

Characteristics of a good explosive.

1. Chemically stable in storage but readily set off by a suitable stimulus.
2. Produces a large volume of gas with high heat of explosion (high brisance/power).
3. High density, definite shattering effect, and safe to handle/transport.

Deflagrating vs detonating.

• Deflagrating: burns/decomposes relatively slowly (subsonic), layer by layer (e.g. gunpowder) — a propellant action.
• Detonating: decomposes through a supersonic shock wave, almost instantaneously throughout the mass (e.g. TNT, RDX) — a shattering (brisant) action.

(a) Ion-exchange (demineralization). The hard water is passed first through a cation-exchange resin ($RH_2$, in H$^+$ form) and then through an anion-exchange resin ($R'(OH)_2$, in OH$^-$ form).

Cation exchange (removes Ca$^{2+}$, Mg$^{2+}$):

$$RH_2 + \text{Ca}^{2+} \rightarrow R\text{Ca} + 2H^+$$ $$RH_2 + \text{Mg}^{2+} \rightarrow R\text{Mg} + 2H^+$$

Anion exchange (removes Cl$^-$, SO$_4^{2-}$, HCO$_3^-$):

$$R'(OH)_2 + \text{SO}_4^{2-} \rightarrow R'\text{SO}_4 + 2OH^-$$

The liberated $H^+$ and $OH^-$ combine to give pure water: $H^+ + OH^- \rightarrow H_2O$.

Regeneration: exhausted cation resin is regenerated with dil. HCl/H$_2$SO$_4$, the anion resin with dil. NaOH:

$$R\text{Ca} + 2HCl \rightarrow RH_2 + \text{CaCl}_2 ; \qquad R'\text{SO}_4 + 2NaOH \rightarrow R'(OH)_2 + \text{Na}_2\text{SO}_4$$

The product is essentially ion-free (demineralized) water.

(b) Disinfection methods (any two):

1. Chlorination — adding chlorine/bleaching powder; forms hypochlorous acid (HOCl) that kills microorganisms.
2. Ozonation — treatment with ozone (O$_3$), a strong oxidising disinfectant.
3. (Also: UV irradiation; boiling.)

(a) Conductance definitions.

• Specific conductance (conductivity, $\kappa$): the conductance of a solution held between two electrodes of unit area (1 cm$^2$) placed unit distance (1 cm) apart, i.e. conductance per unit length per unit cross-section. Unit: S cm$^{-1}$ (SI: S m$^{-1}$).
• Molar conductance ($\Lambda_m$): the conductance of all the ions produced by one mole of electrolyte when placed between electrodes 1 cm apart: $\Lambda_m = \dfrac{\kappa \times 1000}{C}$. Unit: S cm$^2$ mol$^{-1}$ (SI: S m$^2$ mol$^{-1}$). (Equivalent conductance uses normality instead of molarity.)

(b) Mass of Al deposited (Faraday's laws).

Charge passed:

$$Q = I\times t = 2\ \text{A} \times (30\times 60)\ \text{s} = 2 \times 1800 = 3600\ \text{C}$$

Cathode reaction: $\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$, so $n = 3$ ; equivalent weight $= 27/3 = 9$ g.

$$m = \frac{E \times Q}{F} = \frac{(27/3)\times 3600}{96500} = \frac{9 \times 3600}{96500} = \frac{32400}{96500}$$ $$\boxed{m = \mathbf{0.3358\ g} \approx 0.336\ g\ \text{of Al}}$$