催化 Enhancing the rate of a chemical reaction without being consumed Heterogeneous catalysis Without Catalyst in a phase different catalyst from the reactants and products Homogeneous catalysis Catalyst in the same phase as 2 the reactants and products Products Enzyme catalysis pecialized proteins
1 1 催化
均相催化 Chiral Mn(lll) complex Molecules in solution Advantages Good control of catalyst t-Bu structure and composition B I-Bu High activity and selectivity can be achieved Enantioselective Epoxidation · Disadvantages: Difficult separation of products oN and catalyst Limited variation in TOF =6.9s-1 Sel= 95% T=0°C temperature Luo et a.J. Catal.287(2012)170-17
2 2 均相催化
异相催化 High surface area solids Ru-based ammonia synthesis catalyst · Advantages: 5nm Easy separation of products from catalyst Good for flow reactors Large variation in temperature and pressure possible Dis sadvantages Complex structure Hansen et al., Science 294, 1508(2001) Limited control of catalyst structure
3 3 异相催化
酶催化 Highly specialized proteins Nitrogenase High activity in many cases gh specificity Controls all chemical processes in living organisms · Homogenous or heterogeneous? Schindelin et al., Nature 387, 370(1997)
4 4 酶催化
催化的应用 Fertilizer Gasoline OIL Pharmaceuticals oe ndclutcom Plastics image mayomedcinedlog cor Image wiped CD
5 5 催化的应用
催化的应用 c10% of US gnp I M employees in US ·~10% of energy use 90% dependent on catalysis Energy-heavy processes dominated by heterogeneous catalysis
6 6 催化的应用
科学问题与挑战 catalyst design for energy transformations Need efficient catalysts (small losses Need stable catalysts Need catalysts made from Earth-abundant elements Products Surface Need design rules (understanding!)
7 7 科学问题与挑战
cO2的电化学还原 Methane formation 0.1 CO,+8(H+e)→CH4+2H2O 1E-3 1E-4 Zn Very negative potential needed 1E-5 Cu Methanol 0.8VⅴsRHE H Pt 0.1 Thermodynamically, +0. 17Vvs 001 rhe Should be sufficient 1E3 ots of room for improvement -16-14-1.2-1.008060402 EⅣ VS RHE Experiments Kuhl. Cave. Abram. Saha. Kibsgaard Jaramillo Energy Environ. Sci. 5, 7050(2012)
8 8 CO2的电化学还原
催化基础 Reaction energy △E=E product reactant ① E Rate constant ① k=ve-Ealkr AE All determined by energy Reaction coordinate of system
9 9 催化基础
Rate constant k Ea/keT Typical values V=1013s Kpt=0.025 eV at 300K 1. Which activation energy g gives k=1 s1at 300 K? △E 8 2. At which temperature will a process with E=2eV Reaction coordinate have k=1s1?
10 10
