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JEE Advanced levelSingle correct optionDifficulty 3 of 3g3208

A reaction $A B C$ has step 1 with forward $E_{a} = 40 kJ mo l^{- 1}$ and reverse $E_{a} = 10 kJ mo l^{- 1}$ , and step 2 with forward $E_{a} = 25 kJ mo l^{- 1}$ and reverse $E_{a} = 70 kJ mo l^{- 1}$ . The overall $Δ H$ and the rate-determining step are

A $Δ H = - 15 kJ mo l^{- 1}$ ; step 2
B $Δ H = + 15 kJ mo l^{- 1}$ ; step 1
C $Δ H = + 65 kJ mo l^{- 1}$ ; step 2
D $Δ H = - 15 kJ mo l^{- 1}$ ; step 1

Answer: A

Δ H_{1} = 40 - 10 = + 30

and

Δ H_{2} = 25 - 70 = - 45

, so

Δ H = - 15 kJ mo l^{- 1}

. Taking A as zero, TS1 is at 40, B at 30 and TS2 at 55 kJ/mol. TS2 is the highest, so step 2 is rate-determining.

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