Q86 · CSIR-NET Chemistry, December 2011
Paper: CSIR-NET December 2011 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Arrhenius Activation · Marks: 2 · Difficulty: Hard
The rate law for one of the mechanism of the pyrolysis of $\mathrm{CH}_{3} \mathrm{CHO}$ at $520^{\circ} \mathrm{C}$ and 0.2 bar is \[ \text { Rate }=-\left|\mathrm{k}_{2}\left(\frac{\mathrm{k}_{1}}{2 \mathrm{k}_{1}}\right)^{1 / 2}\right|\left[\mathrm{CH}_{3} \mathrm{CHO}\right]^{3 / 2} \]
The overall activation energy E, in terms of the rate law is: (a)$\mathrm{Ea}(2)+\mathrm{Ea}(1)+2 \mathrm{Ea}(4)$
(b)$\mathrm{Ea}(2)+\frac{1}{2} \mathrm{Ea}(1)-\mathrm{Ea}(4)$
(c)$\mathrm{Ea}(2)+\frac{1}{2} \mathrm{Ea}(1)-\frac{1}{2} \mathrm{Ea}(4)$
(d)$\mathrm{Ea}(2)-\frac{1}{2} \mathrm{Ea}(1)-\frac{1}{2} \mathrm{Ea}(4)$
Answer
Answer: B ✓ checked by 4AB · confidence medium
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
Rice–Herzfeld: k = k₂(k₁/2k₄)^{1/2}, so E_a = E_a(2) + ½E_a(1) − ½E_a(4).
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