Question

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A first order gas phase reaction : A₂B₂(g) → 2A(g) + 2B(g) at the temperature 400°C has the rate constant k = 2.0 × 10⁻⁴ sec⁻¹. What percentage of A₂B₂ is decomposed on heating for 900 seconds? (Antilog 0.0781 = 1.197)

Answer 1

Solution :-

$:\implies\sf k = \frac{2.303}{t} \: log \frac{9}{9 - x} \\ \\ \\ :\implies\sf \: 2 \times {10}^{ - 4} = \frac{2.303}{900} log \frac{100}{100 - x}$

$:\implies\sf \: log \frac{100}{100 - x} = \frac{9}{115.15} = 0.0781\\ \\ \\ :\implies\sf \: \frac{100}{100 - x} = antilog(0.0781) \\ \\ \\ :\implies\sf \: 100 = 1.197(100 - x)$

$:\implies\sf 1.119.7 - 1.197x \\ \\ \\:\implies\sf 1.197x = 19.7 \\ \\ \\ :\implies\sf x = \cancel \frac{19.7}{1.197} \\ \\ \\ :\implies\sf \: x = 16.45\%$

so therefore our required answer is 16.45%

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Answer 2

꧁Hope it helps uhh ꧂

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