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whats the answer to this questions

science chemistry

Answer 1

Answer #1

From the given data

Part 1)

activation energy = -slope x 8.314

= -(-122.32) x 8.314

= 1017 J/mol

Part 2)

k at T1 = 300 K

k2 = 2.14 x 10^5 at T2 = 658 K

Using,

ln(k2/k1) = Ea/R[1/T1 - 1/T2]

ln(2.14 x 10^5/k1) = 1017/8.314[1/300 - 1/658]

k1 = 1.71 x 10^5 M-1.s-1

Answer 2

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