JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
A sample of milk splits after \(60 \;min\). at \(300\; \mathrm{K}\) and after \(40 \;min\). at \(400 \;\mathrm{K}\) when the population of lactobacillus acidophilus in it doubles. The activation energy (in \(\mathrm{kJ} / \mathrm{mol}\) ) for this process is closest to ............. \(\mathrm{kJ/mole}\) (Given, \(\mathrm{R}=8.3\; \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}\)\(, \ln \left(\frac{2}{3}\right)=0.4\) \(\left.e^{-3}=4.0\right)\)
- A \(2.88\)
- B \(2.52\)
- C \(1.96\)
- D \(3.98\)
Answer & Solution
Correct Answer
(D) \(3.98\)
Step-by-step Solution
Detailed explanation
\(\ln \left(\frac{t_{1}}{t_{2}}\right)=\frac{-\operatorname{Ea}}{R}\left[\frac{1}{T_{2}}-\frac{1}{T_{1}}\right]\) \(\ln \left(\frac{60}{40}\right)=\frac{-\operatorname{Ea}}{8.3}\left[\frac{1}{400}-\frac{1}{300}\right]\) \(\mathrm{E}=0.4 \times 1200 \times 8.3\)…
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