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JEE MainChemistryToughChemical Kinetics

The reaction A(s) → 2 B(g) + C(g) is first order. The pressure
after 20 min and after very long time are 150 mm Hg and 225 mm Hg. The
value of rate constant and pressure after 40 min are

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A(s) → 2 B(g) + C(g)
K=1tIn(P∞−P0P∞−Pt)        P0=0              Pt=150                         P∞=225K = \frac{1}{t}{In}\left( \frac{P_{\infty} - P_{0}}{P_{\infty} - P_{t}} \right){\ \ \ \ \ \ \ \ P}_{0} = 0\ \ \ {\ \ \ \ \ \ \ \ \ \ \ P}_{t} = 150\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P}_{\infty} = 225
120In[225225−150]⇒120In(22575)=0.05ln3\frac{1}{20}In\lbrack\frac{225}{225 - 150}\rbrack \Rightarrow \frac{1}{20}In(\frac{225}{75}) = 0.05ln3
120ln(22575)=140ln⁡(225225×P40)⇒(22575)2=225225×P40\frac{1}{20}ln(\frac{225}{75}) = \frac{1}{40}\ln\left( \frac{225}{225 \times P_{40}} \right) \Rightarrow {(\frac{225}{75})}^{2} = \frac{225}{225 \times P_{40}}
22575×75=1225−P40⇒P40=200\frac{225}{75 \times 75} = \frac{1}{225 - P_{40}} \Rightarrow P_{40} = 200

Source: Grass-G / CH-15 Chemistry Paper 11 Nov..docx

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