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AITS & Test SerieshardNUMERICAL

Question

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Answer: 00002.00

💡 Solution & Explanation

f b a a H E E    b a E = 120 kJ mol–1 After addition of catalyst H = -40 kJ mol–1 fa E = 20 kJ mol–1 b a E = 60 kJ mol–1

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