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Grade 11IBChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Materials can be classified into four main categories: metals, polymers, ceramics, and composites, based on their structures and properties.

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Metals are characterized by a 'sea' of delocalized electrons. The properties of metals can be modified by alloying, where atoms of different sizes disrupt the regular lattice, making it harder for layers to slide (increasing hardness).

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Inductively Coupled Plasma (ICP) spectroscopy is used to determine trace metal concentrations. It uses a high-temperature plasma (60006000 KK to 1000010000 KK) to atomize and excite samples, which then emit light at characteristic wavelengths.

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Catalysts lower the activation energy (EaE_a) of a reaction by providing an alternative reaction pathway. Heterogeneous catalysts provide a surface for reactants to adsorb, while homogeneous catalysts are in the same phase as the reactants.

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Liquid crystals are substances that exhibit a phase between a solid and a liquid. Thermotropic liquid crystals change phase depending on temperature, and their molecular orientation can be controlled by an electric field, which is the basis for LCD technology.

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Polymers are formed via addition or condensation reactions. Their properties depend on the degree of branching, chain length, and cross-linking (e.g., S−SS-S bridges in vulcanized rubber).

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Nanotechnology involves particles sized between 11 nmnm and 100100 nmnm. These materials have high surface area to volume ratios (SA:VSA:V), leading to unique chemical and physical properties compared to bulk materials.

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Superconductors are materials that have zero electrical resistance below a critical temperature (TcT_c). Type I superconductors have a sharp transition, while Type II superconductors have a gradual transition and can operate at higher magnetic fields.

📐Formulae

nλ=2dsin⁡θn\lambda = 2d \sin \theta

Atom Economy=Molar mass of desired productTotal molar mass of all reactants×100%\text{Atom Economy} = \frac{\text{Molar mass of desired product}}{\text{Total molar mass of all reactants}} \times 100\%

SA:V Ratio (Cube)=6L2L3=6L\text{SA:V Ratio (Cube)} = \frac{6L^2}{L^3} = \frac{6}{L}

ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S

Efficiency=Useful energy outputTotal energy input×100%\text{Efficiency} = \frac{\text{Useful energy output}}{\text{Total energy input}} \times 100\%

💡Examples

Problem 1:

Calculate the atom economy for the production of Iron (FeFe) in the thermite reaction: Fe2O3(s)+2Al(s)→Al2O3(s)+2Fe(l)Fe_2O_3(s) + 2Al(s) \rightarrow Al_2O_3(s) + 2Fe(l). (Molar masses: Fe2O3=159.70Fe_2O_3 = 159.70, Al=26.98Al = 26.98, Fe=55.85Fe = 55.85, Al2O3=101.96Al_2O_3 = 101.96 g/molg/mol)

Solution:

Total mass of reactants=159.70+2(26.98)=213.66 g/mol\text{Total mass of reactants} = 159.70 + 2(26.98) = 213.66 \text{ g/mol} Mass of desired product (2Fe)=2×55.85=111.70 g/mol\text{Mass of desired product (2Fe)} = 2 \times 55.85 = 111.70 \text{ g/mol} Atom Economy=111.70213.66×100%≈52.28%\text{Atom Economy} = \frac{111.70}{213.66} \times 100\% \approx 52.28\%

Explanation:

Atom economy measures the proportion of reactant atoms that end up in the desired product. Higher values indicate more 'green' or sustainable chemical processes.

Problem 2:

X-rays of wavelength 0.1540.154 nmnm are directed at a crystal. The first-order reflection (n=1n=1) occurs at an angle θ=15.5∘\theta = 15.5^{\circ}. Calculate the interplanar distance dd.

Solution:

Using Bragg's Law: nλ=2dsin⁡θn\lambda = 2d \sin \theta 1×0.154=2×d×sin⁡(15.5∘)1 \times 0.154 = 2 \times d \times \sin(15.5^{\circ}) d=0.1542×0.2672≈0.288 nmd = \frac{0.154}{2 \times 0.2672} \approx 0.288 \text{ nm}

Explanation:

Bragg's Law relates the wavelength of electromagnetic radiation to the diffraction angle and the lattice spacing in a crystalline sample.

Problem 3:

Compare the surface area to volume ratio of a cubic nanoparticle with side length 1010 nmnm to a bulk cube with side length 11 cmcm.

Solution:

For the 1010 nmnm cube: SA:V=610×10−9=6×108 m−1SA:V = \frac{6}{10 \times 10^{-9}} = 6 \times 10^8 \text{ m}^{-1} For the 11 cmcm (0.010.01 mm) cube: SA:V=60.01=600 m−1SA:V = \frac{6}{0.01} = 600 \text{ m}^{-1}

Explanation:

The nanoparticle has a significantly higher SA:VSA:V ratio (10610^6 times greater), explaining why nanomaterials are often highly reactive catalysts.