Question

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?

Answer 1

a) 7.96* 10⁻³ s

b) 0.0112 s

c) 1.645* 10⁻³⁸ g

Step-by-step 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