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The thermal decomposition of N₂O₅(g) to form NO₂(g) and O₂(g) is a first-order reaction. The rate constant for the reaction is 5.1 × 10⁻⁴ s⁻¹ at 318 K. What is the half-life of this process?

a) 1354 seconds
b) 2153 seconds
c) 1562 seconds
d) 2279 seconds

User DBlisse
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1 Answer

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Final answer:

The half-life of the thermal decomposition of N₂O₅(g) to form NO₂(g) and O₂(g) is approximately 1354 seconds.

The half-life of the first-order thermal decomposition of N2O5(g) with a rate constant of 5.1 × 10⁻⁴ s⁻¹ is approximately 1354 seconds, calculated using the formula t_{1/2} = 0.693/k.

Step-by-step explanation:

The thermal decomposition of N₂O₅(g) to form NO₂(g) and O₂(g) is a first-order reaction. The rate constant for the reaction is given as 5.1 × 10⁻⁴ s⁻¹ at 318 K. The half-life of a first-order reaction can be calculated using the formula:

t1/2 = (0.693 / k)

Plugging in the given rate constant, the half-life can be calculated as:

t1/2 = (0.693 / 5.1 × 10⁻⁴ s⁻¹) ≈ 1354 seconds

The half-life of the first-order thermal decomposition of N2O5(g) with a rate constant of 5.1 × 10⁻⁴ s⁻¹ is approximately 1354 seconds, calculated using the formula t_{1/2} = 0.693/k.

The half-life for a first-order reaction is given by the formula t_{1/2} = \frac{0.693}{k}, where k is the rate constant. Given the rate constant for the thermal decomposition of N2O5(g) is 5.1 \times 10^{-4} s^{-1} at 318 K, we can calculate the half-life as follows:



t_{1/2} = \frac{0.693}{5.1 \times 10^{-4} s^{-1}} = \frac{0.693}{0.00051 s^{-1}} \approx 1358.82 seconds.



Therefore, the correct answer would be closest to 1354 seconds (option a).

User Akrikos
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