Consider the reaction 3A + B + C ? D + E where the rate law is defined as Soluti

Question

Consider the reaction 3A + B + C ? D + E where the rate law is defined as

Explanation / Answer

Since B and C are present in large excess the concentration change due to reaction is small So you can approximate their concentration to be constant at initial level. [B] ˜ [B]0 [C] ˜ [C]0 The rate law reduces to a pseudo second order rate law -d[A]/dt = k·[B]0·[C]0·[A]² => -1/[A]² d[A] = k·[B]0·[C]0 dt ? 1/[A] d[A] = ? k·[B]0·[C]0 dt 1/[A] = k·[B]0·[C]0·t + c apply initial condition to find c [A]t=0 = [A]0 1/[A]0 = k·[B]0·[C]0·t + c c = 1/[A]0 Hence 1/[A] = k·[B]0·[C]0·t + 1/[A]0 [A] = [A]0 / (k·[A]0·[B]0·[C]0·t + 1) solve for k and insert the values measured after 3min k = (1/[A] - 1/[A]0 ) / ([B]0·[C]0·t ) = (1/3.26×10?5M - 1/1.0×10?4M ) / (1.0M · 1.0M · 3.0min ) = 6891.6M?³min?¹ (b) at half life t½ concentration has dropped to [A]0/2 Hence: [A]0/2 = [A]0 / (k·[A]0·[B]0·[C]0·t½ + 1) 2 = k·[A]0·[B]0·[C]0·t½ + 1 t½ = 1/(k·[A]0·[B]0·[C]0 ) = 1/( 6891.6M?³min?¹ · 1.0×10?4M · 1M · 1M ) = 0.6892min = 41.3s

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