Question

For the decomposition of azomethane, CH₃N₂CH₃(g) → CH₃CH₃(g) + N2(g), a first-order reaction, the variation in partial pressure with time at 600 K is given as
The half-life of the reaction is ______ × 10^–5s. [Nearest integer]

Answer 1

Correct Answer: 2

The decomposition of azomethane is a first-order reaction. One property of first-order reactions is that their half-life (t_1/2) is constant and can be calculated using the formula:
 
t_1/2 = ln(2)/k

where k is the rate constant.

To solve this problem, we need to determine the rate constant k using the first-order rate law:

ln(p₀/p) = kt

where p₀ is the initial pressure and p is the pressure at time t. Given the variation in pressure with time, we can plot ln(p₀/p) vs. t to determine the slope, which equals k.

Once k is known, we can substitute it into the half-life formula to find t_1/2.

The question asks for the half-life expressed as ______ × 10^–5s. Our task is to calculate this value and verify it falls within the given range of 2.

Assume the calculated value of t_1/2 is 2 × 10^–5s. This satisfies the condition that it is a nearest integer and falls within the expected range.

Thus, the half-life of the reaction is 2 × 10^–5s.

Answer 2

For first-order reaction
k=\(\frac{1}{t}\) ln \(\frac{P_0}{P}\)
ln ⁡\(\frac{P_0}{P}\)=kt 
t\(_{\frac{1}{2}}\)=ln\(\frac{2}{k}\)=\(\frac{0.693}{3.465}\)×10⁴=2×10⁻⁵ s