What is the activation energy for the decomposition of N2O5 as N2O5⇌2NO2+12O2If the values of the rate constants are 3.45×10−3 and 6.9×10−3 at 27oC and 67oC respectively.

log⁡6.9×10−33.45×10−3=Ea(340−300)8.314×10−3×2.303×300×340∴ Ea=0.301×8.314×10−3×2.303×300×34040 =14.7 KJ

What is the activation energy for the decomposition of N2O5 as N2O5⇌2NO2+12O2If the values of the rate constants are 3.45×10−3 and 6.9×10−3 at 27oC and 67oC respectively.

log⁡6.9×10−33.45×10−3=Ea(340−300)8.314×10−3×2.303×300×340∴ Ea=0.301×8.314×10−3×2.303×300×34040 =14.7 KJ

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What is the activation energy for the decomposition of $N_{2} O_{5}$ as
$N_{2} O_{5} ⇌ 2 {NO}_{2} + \frac{1}{2} O_{2}$
If the values of the rate constants are $3 .45 \times 10^{- 3}$ and $6 .9 \times 10^{- 3} {at 27}^{o} C {and 67}^{o} C$ respectively.

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$102 \times 10^{2} kJ$

$488.5 kJ$

$112 kJ$

$14.7 kJ$

answer is D.

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Detailed Solution

$\begin{array}{l} \log \frac{6 .9 \times 10^{- 3}}{3 .45 \times 10^{- 3}} = \frac{E_{a} (340 - 300)}{8 .314 \times 10^{- 3} \times 2 .303 \times 300 \times 340} \\ ∴ E_{a} = \frac{0 .301 \times 8 .314 \times 10^{- 3} \times 2 .303 \times 300 \times 340}{40} \\ = 14 .7 KJ \end{array}$