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Normal Reaction

The energy of...

Question

The energy of activation for forward and backward reaction $(E_{f} a n d E_{b})$ at 300 K, Given that $E_{f}$ and $E_{b}$ are in the ratio 20 : 31.

69, 93

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60, 93

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93, 60

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90, 60

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Solution

The correct option is B $60, 93$
$\frac{E_{f}}{E_{b}} = \frac{20}{31}$ (given). Also $E_{f} - E_{b} = - 33$ (given) ..... (i)
Substituting $E_{b} = \frac{31}{20} E_{f}$ , in equation (i) we get
$E_{f} - \frac{31}{20} E_{f} = - 33$
or $\frac{- 11}{20} E_{f} = - 33$ or $E_{f} = \frac{33 \times 20}{11} = 60 k J m o l^{- 1}$
$E_{b} = E_{f} + 33 = 60 + 33 = 93 k J m o l^{- 1}$

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The energy of activation for forward and backward reaction (EfandEb) at 300 K, Given that Ef and Eb are in the ratio 20 : 31.

The correct option is B 60,93EfEb=2031 (given). Also Ef−Eb=−33 (given) ..... (i)Substituting Eb=3120Ef, in equation (i) we getEf−3120Ef=−33or −1120Ef=−33 or Ef=33×2011=60kJmol−1Eb=Ef+33=60+33=93kJmol−1

The energy of activation for forward and backward reaction $(E_{f} a n d E_{b})$ at 300 K, Given that $E_{f}$ and $E_{b}$ are in the ratio 20 : 31.