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Standard XII

Chemistry

Polymerization Isomerism

The rate cons...

Question

The rate constant for two parallel reactions were found to be $1.0 \times 10^{- 2} d m^{3}$ $m o l^{- 1} s^{- 1}$ and $3.0 \times 10^{- 2} d m^{3}$ $m o l^{- 1} s^{- 1}$ . If the corresponding energies of activation of the parallel reactions are 60.0 KJ $m o l^{- 1}$ and 70.0 KJ $m o l^{- 1}$ respectively, then what is the apparent overall energy of activation?

130.0 KJ

m o l^{- 1}

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65.0 KJ

m o l^{- 1}

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67.5 KJ

m o l^{- 1}

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100.0 KJ

m o l^{- 1}

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Solution

The correct option is C 67.5 KJ $m o l^{- 1}$
Overall energy of activation = $\frac{k_{1} E_{1} + k_{2} E_{2}}{k_{1} + k_{2}}$
$⟹ \frac{60 \times 1 \times 10^{- 2} + 70 \times 3 \times 10^{- 2}}{4 \times 10^{- 2}}$
$⟹ \frac{270}{4} = 67.5 K J m o l^{- 1}$

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Similar questions

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The correct option is C 67.5 KJ mol−1Overall energy of activation =k1E1+k2E2k1+k2⟹60×1×10−2+70×3×10−24×10−2⟹2704=67.5KJmol−1

The correct option is C 67.5 KJ $m o l^{- 1}$
Overall energy of activation = $\frac{k_{1} E_{1} + k_{2} E_{2}}{k_{1} + k_{2}}$
$⟹ \frac{60 \times 1 \times 10^{- 2} + 70 \times 3 \times 10^{- 2}}{4 \times 10^{- 2}}$
$⟹ \frac{270}{4} = 67.5 K J m o l^{- 1}$