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 __________ × 10⁻⁵ s. [Nearest integer]

Solution

Chemistry ma’am · 4 steps

For a first order plot of ln(p/p0) against t, the slope is just −k. Read k off the graph.

1. First order law. For a first order gas reaction, the partial pressure of azomethane falls as p = p₀ e⁻ᵏᵗ. Taking the natural log gives a straight line through the origin.

   ln(p/p₀) = -kt

2. Read the slope. The graph plots ln(p/p₀) against t, so its slope equals -k. The slope is -3.465 × 10⁴, so k = 3.465 × 10⁴ s⁻¹.
3. Half-life. t_(1/2) = 0.693/k = 0.693/(3.465 × 10⁴) = 0.2 × 10⁻⁴ s = 2 × 10⁻⁵ s.
4. So. The half-life is 2 × 10⁻⁵ s, so the answer is 2.

The common slip: 1. Treating the plot as a log₁₀ plot makes k = 2.303 × 3.465 × 10⁴, which gives t_(1/2) ≈ 0.87 × 10⁻⁵ s and rounds to 1, not 2.

The idea to keep. Always check the axes of a first order plot. With ln on the y-axis the slope is −k. With log10 it is −k/2.303. Then use t½ = 0.693/k.

Answer from the NTA final answer key. Solution by PrepWiser.