Question

A first-order reaction is 50% completed in 1.26 x 10¹⁴s. How much time would it take for 100% completion?

1.26 x 10¹⁵ s

2.52 x 10¹⁴ s

2.52 x 10²⁸ s

Infinite

Solution

Infinite

K- for the first order,

$\frac{0.693}{t_{1 / 2}} = \frac{0.693}{1.26 x 10^{14}} for 100 % completion ∵ t = \frac{2.303}{k} \log \frac{a}{a - x} t = \frac{2.303 x 1.26 x 10^{14}}{0.693} \log \frac{100}{0} t = \infty$

Some More Questions From Chemical Kinetics Chapter

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Chemical Kinetics

A first-order reaction is 50% completed in 1.26 x 10¹⁴s. How much time would it take for 100% completion?

1.26 x 10¹⁵ s

2.52 x 10¹⁴ s

2.52 x 10²⁸ s

Infinite

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A reaction is 50% complete in 2 hours and 75% complete in 4 hours. What is the order of the reaction.

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Chemical Kinetics

A first-order reaction is 50% completed in 1.26 x 1014s. How much time would it take for 100% completion? 1.26 x 1015 s 2.52 x 1014 s 2.52 x 1028 s Infinite

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What will be the effect of temperature on rate constant?

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Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

A first-order reaction is 50% completed in 1.26 x 10¹⁴s. How much time would it take for 100% completion?

1.26 x 10¹⁵ s

2.52 x 10¹⁴ s

2.52 x 10²⁸ s

Infinite

The rate law for the decomposition of N₂O₅ is rate = k[N₂O₅]. What is the significance of k in this equation?