The first order rate constant for the conversion of cyclobutane to ethylene at 1000.°C is 87 1/s. Cyclobutane â 2 ethylene (a) what is the half-life of this reaction at 1000.°C (b) If you started with 4.00 g of cyclobutane, how long would it take to consume 2.50 g of it? (c) How much of an initial 1.00 g sample of cyclobutane would remain after 1.00 s?

a) 7.96 * 10⁻³ s

b) 0.0112 s

c) 1.645 * 10⁻³⁸ g

Explanation:

for the reaction

Cyclobutane (A) → 2 ethylene (B)

the reaction rate (first order) is

-dCa/dt = k*Ca

∫dCa/Ca = - ∫k*dt

ln (Ca/Ca₀) = - k*t → Ca = Ca₀*e^ (-k*t)

therefore

a) the half - life represents the time required for the concentration Ca to drop to half of the initial value (Ca=Ca₀/2) therefore

Ca₀/2 = Ca₀*e^ (-k*t) → - ln 2 = - k*t → t = ln 2 / k = ln 2 / (87 1/s) = 7.96 * 10⁻³ s

b) Ca = Ca₀*e^ (-k*t), for Ca = Wa / (V*M), where Wa is weight:

Wa = Wa₀*e^ (-k*t)

for Wa₀ = 4 g and Wa = 4g - 2.5 g = 1.5 g

→ t = (1/k) * ln (Wa₀/Wa) = 1 / (87 1/s) * ln [ 4g / (1.5 g) ] = 0.0112 s

c) for Wa₀ = 4 g and t=1 s

Wa = Wa₀*e^ (-k*t) = 1 g * e^ ( - 87 (1/s) * 1 s) = 1.645 * 10⁻³⁸ g ≈ 0