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Standard XII

Chemistry

Half life

^M_2A(g)→ ^M-...

Question

$_{2}^{M} A (g) \to_{Z - 4}^{M - 8} B (g) + α - p a r t i c l e s$
The radioactive disintegration follows first order kinetics. Starting with 1 mol of in a 1 litre closed flask at $27^{\circ}$ C, pressure set up after two half-lives is approximately

25 atm

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12 atm

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37 atm

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40 atm

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Solution

The correct option is C 37 atm
$I n i t i a l P = \frac{1 \times 0.0821 \times 300}{1} = 24.63_{2}^{M} A (g) \to_{Z - 4}^{M - 8} B (g) + α - p a r t i c l e s 100 1 - 0.75 0.75 2 \times 0.75 = 1.5$
= 0.25
Total no.of moles = 0.25 + 0.75 + 1.5 = 2.5
$\frac{P_{f}}{P_{i}} = \frac{2.5}{1} P_{f} = 2.5 \times 24.63$
Pressure increased = 61.57-24.63
= 37 atm

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Similar questions

_{2}^{M} A (g) \to_{Z - 4}^{M - 8} B (g) + α - p a r t i c l e s

The radioactive disintegration follows first order kinetics. Starting with 1 mol of in a 1 litre closed flask at

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C, pressure set up after two half-lives is approximately

_{2}^{M} A (g) \to_{Z - 4}^{M - 8} B (g) + α - p a r t i c l e s

The radioactive disintegration follows first order kinetics. Starting with 1 mol of in a 1 litre closed flask at

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M2A(g)→M−8Z−4B(g)+α−particles The radioactive disintegration follows first order kinetics. Starting with 1 mol of in a 1 litre closed flask at 27∘C, pressure set up after two half-lives is approximately

The correct option is C 37 atmInitial P=1×0.0821×3001=24.63M2A(g)→M−8Z−4B(g)+α−particles1 0 01−0.75 0.75 2×0.75=1.5 = 0.25 Total no.of moles = 0.25 + 0.75 + 1.5 = 2.5 PfPi=2.51Pf=2.5×24.63 Pressure increased = 61.57-24.63 = 37 atm

$_{2}^{M} A (g) \to_{Z - 4}^{M - 8} B (g) + α - p a r t i c l e s$
The radioactive disintegration follows first order kinetics. Starting with 1 mol of in a 1 litre closed flask at $27^{\circ}$ C, pressure set up after two half-lives is approximately