Answer is: A) 124 s. c₀ = 3 mol/L. c₁ = 0,700 mol/L. k = 8,8·10⁻³ 1/M·s. Integrated second order rate law is: 1/c₁ = 1/c₀ + k·t. k·t = 1/0,700 - 1/3. 0,0088·t = 1,095. t = 1,095 ÷ 0,0088. t = 124 s. c₀ - <span>initial concentration. c</span>₁ - <span> concentration at a particular time. k - </span><span>the rate constant. t - time.</span>