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5. How much faster would a reaction be if a catalyst is used that lowers the activation energy from 20.0 kJ/mol to 10.0 kJ/mo
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Answer #1

Ea= activation energy * using Arrehenius equation -EalRT K= A e at T=26°c K at 20 kJ/mol K2 at 10 kJ/mol P=128+273= 298K -20xK2 0.012206 KI 1.489x104 87.93 Reaction is 87.93 times faster wun catalyst is used at 0°C

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