A hydrogenation reaction is carried out at 500 K. CH2 = CH2 + H2 xrightarrow[ textno catalyst]500 K CH3 - CH3 Activation energy - Ea kJ mol -1 CH2 = CH2 + H2 xrightarrowPd. 400 K CH3 - CH3 Activation energy = (Ea - 20) kJ mol -1 If rate remains constant, then Ea is

Question

A hydrogenation reaction is carried out at 500 K. CH2 = CH2 + H2 xrightarrow[ textno catalyst]500 K CH3 - CH3 Activation energy - Ea kJ mol -1 CH2 = CH2 + H2 xrightarrowPd. 400 K CH3 - CH3 Activation energy = (Ea - 20) kJ mol -1 If rate remains constant, then Ea is (A) 120 kJ mol -1 (B) 100 kJ mol -1 (C) 20 kJ mol -1 (D) 80 kJ mol -1

Tardigrade Admin · Accepted Answer

k = Ae-Ea/RT k =Ae-Ea/R ×500 k' = Ae-(Ea=20)/R×400 Rates are equal, hence ∴ Ae-Ea/R ×500 =Ae-(Ea-20)R ×400 ∴ (Ea/500) = (Ea -20/400) 4Ea = 5Ea -100 ∴ Ea = 100 kJ mol -1

$120 k J$ mol $^{- 1}$

$100 k J$ mol $^{- 1}$

$20 k J$ mol $^{- 1}$

$80 k J$ mol $^{- 1}$

Q. A hydrogenation reaction is carried out at 500K. CH2​=CH2​+H2​500Kno catalyst​CH3​−CH3​ Activation energy −Ea​kJ mol −1 CH2​=CH2​+H2​Pd.400K​CH3​−CH3​ Activation energy =(Ea​−20)kJ mol −1 If rate remains constant, then Ea​ is

120kJ mol −1

100kJ mol −1

20kJ mol −1

80kJ mol −1

k=Ae−Ea​/RT k=Ae−Ea​/R×500 k′=Ae−(Ea​=20)/R×400 Rates are equal, hence ∴Ae−Ea​/R×500=Ae−(Ea​−20)R×400 ∴500Ea​​=400Ea​−20​ 4Ea​=5Ea​−100 ∴Ea​=100kJ mol −1

$120 k J$ mol $^{- 1}$

$100 k J$ mol $^{- 1}$

$20 k J$ mol $^{- 1}$

$80 k J$ mol $^{- 1}$