(a) A new photocatalyst with a surface area of 2 m2 has recently been developed, which splits water using the Sun's UV l

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A A New Photocatalyst With A Surface Area Of 2 M2 Has Recently Been Developed Which Splits Water Using The Sun S Uv L 1 (119.52 KiB) Viewed 65 times

(a) A new photocatalyst with a surface area of 2 m2 has recently been developed, which splits water using the Sun's UV light at a quantum yield of 60%. The photocatalyst can absorb 50% of UV light reaching it; 3% of sunlight is UV. (i) On May 15, 2021 (Day Number 132), calculate the number of hours of sunlight that this photocatalyst could be exposed to, assuming clear sky conditions. (4 marks) (ii) Using the relevant formulae in the APPENDIX, calculate the mass of hydrogen gas that can be produced by the photocatalyst, when exposed for the number of hours that you obtained in Q3a(i). (4 marks) (iii) Calculate the overall efficiency of the photocatalysis process if the hydrogen gas produced is combusted in an engine at 80% efficiency. Take the enthalpy of combustion of hydrogen gas as -285.5 kJ mol-1. (2 marks)
Ohm's Law, V = IR Charge: 1 e = 1.60*10-19 C Current: 1 A = 1 Cs Voltage: 1 V = 1 JC1 n Quantum Yield, nph Wein's Displacement Law; Amax = b/T (T in K and b = 2.9 x10mk) Volume of a sphere = (4/3) Ter? Surface area of a sphere = 4 Ter2 Surface area of the Sun = 6.09x1012 km2 Conversion Efficiency Energyout Energyin Stefan-Boltzmann's constant, 0 = 5.7x108 W m2K Avogadro's constant, NA = 6.02x1023 mol-1 Planck's Constant, h = 6.23x10-34Js Energy: 1 eV = 1.60x10-19 J Speed of light, c = 3.0x108 ms Photon Energy, E = 1243/2 (E in eV) AM1.5 = Air Mass Coefficient universally applicable for characterising terrestrial power-generating panels = 1000 W m2 Enthalpy Change of Combustion Data AHC (H2) = -285.8 kJ mol-1 Enthalpy Change of Formation Data AHÍ (CO2) = -393.5 kJ mol-1 AHÍ (CO) = -110.5 kJ mol-1 AHF (CHA) = -74.9 kJ mol AHf (H2O) = -285.8 kJ mo1 Specific heat capacity of water is 4200 J kg K1 1 toe = 41.87 Giga Joules (GJ) 1 kWh m2 day1 = 41.67 W m2 1 kWh y = 0.114 W 1 kWh = 3.6 MJ Band gaps (Eg) of semiconductors Wavelengths, 1 Ge = 0.67 eV Si = 1.12 eV GaAs = 1.43 eV AloGao As = 1.92 eV Gap = 2.23 eV Infra-Red = 750 - 1000000 nm Red = 620 - 750 nm Orange = 590 - 620 nm Yellow = 570-590 nm Green = 495 - 570 nm
InoGao. N = 2.82 eV Zno = 3.37 eV TiO2 = 3.37 eV C (diamond) = 5.50 eV Blue = 450-495 nm Violet = 380- 450 nm UV = 10 - 400 nm For UV calculations, 1 = 380 nm = Solar Angles t-12 Definitions w 24 Hour angle, w= (2) 360° t = time (hours) based on 24-hour clock Zenith angle, Oz cos 8z = sind sind + cosi coso cosw Angle of declination, 8 -23.45 cos * 3600) 0 = latitude B = angle of inclination 0 = 52.5°N for Birmingham, UK D = Day number D = 172 for 2019 Summer solstice D = 356 for 2019 Winter solstice D+10 365 Angle of incidence, cos 0 = cos cos(" -"B) + sin 8 sin(" - "B) Semiconductor properties Reverse Saturation Current Equation: n = AT32 e-Eg/2KT l. = BT e-Eg/T A = 2.8 1022 m3K-3/2 for silicon k = Boltzmann's constant = 8.625 x 10-5 eV K-1 B=0.2 A K3 for silicon Diode Equation: 1; = 1.(VKT-1) Iph = 1 + Ipv T = temperature (K) Eg = Band gap (V) n = electron density (m) in conduction band Iph = photocurrent Ipv = current output from PV cell 1} = current consumed by diode