For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained. t (sec): 0, 360, 720 P(mm of Hg): 35.0, 54.0, 63.0 Calculate the rate constant.
The rate constant for the decomposition of azoisopropane is ⁻³ s⁻¹.
Explanation
This is a first-order gas phase decomposition reaction. For a reaction where one mole of reactant produces two moles of products (hexane and nitrogen), the pressure relationship can be derived. The initial pressure of azoisopropane decreases as it decomposes, while the total pressure increases due to formation of products. Using the first-order integrated rate law with pressure terms, we can calculate k from the given data at different time intervals.
Solution Steps
Step 1: Write the reaction stoichiometry Azoisopropane(g) → Hexane(g) + N₂(g) Let initial pressure = p₀ = 35.0 mm Hg At time t, if x mm Hg decomposes, then:
- Pressure of azoisopropane remaining = p₀ - x
- Pressure of products formed = x + x = 2x
- Total pressure P_t = (p₀ - x) + 2x = p₀ + x
Step 2: Derive pressure of reactant at time t From P_t = p₀ + x, we get x = P_t - p₀ Pressure of azoisopropane at time t = p₀ - x = p₀ - (P_t - p₀) = 2p₀ - P_t
Step 3: Calculate k at t = 360 sec P_t = 54.0 mm Hg Pressure of azoisopropane = 2(35.0) - 54.0 = 16.0 mm Hg k = (2.303/t) × log(p₀/p_A) k = (2.303/360) × log(35.0/16.0) k = (2.303/360) × log(2.1875) k = (2.303/360) × 0.340 = ⁻³ s⁻¹
Step 4: Calculate k at t = 720 sec P_t = 63.0 mm Hg Pressure of azoisopropane = 2(35.0) - 63.0 = 7.0 mm Hg k = (2.303/720) × log(35.0/7.0) k = (2.303/720) × log(5.0) k = (2.303/720) × 0.699 = ⁻³ s⁻¹
Step 5: Average rate constant k_avg = (2.18 + 2.24)/⁻³ = ⁻³ s⁻¹