Q.26
(i) Draw the structure of 2, 4 – dinitrophenylhydrazone of acetaldehyde.
(ii) The rate of A + B → P is 3.6 × 10−2 mol/dm3/s. When [A] = 0.2 M and [B] = 0.1 M. Calculate the rate constant if reaction is first order in B and second order in A.
Best answer
Answer: (i) CH3–CH=N–NH–C6H3(NO2)2 (ii) \(k = 9\) dm\(^6\) mol\(^{-2}\) s\(^{-1}\)
(i) Acetaldehyde condenses with 2,4-dinitrophenylhydrazine (2,4-DNP) with loss of water:
CH3CHO + H2N–NH–C6H3(NO2)2 → CH3CH=N–NH–C6H3(NO2)2 + H2O
Structure (described): CH3–CH=N–NH– attached to C-1 of a benzene ring; the ring carries –NO2 groups at C-2 (ortho to the NH) and C-4 (para to the NH). It is a yellow-orange crystalline solid (acetaldehyde 2,4-dinitrophenylhydrazone).
(ii) Rate law: rate \(= k[\text{A}]^2[\text{B}]\).
\[k = \frac{\text{rate}}{[\text{A}]^2[\text{B}]} = \frac{3.6\times10^{-2}}{(0.2)^2\times0.1} = \frac{3.6\times10^{-2}}{4\times10^{-3}} = 9\ \text{dm}^6\ \text{mol}^{-2}\ \text{s}^{-1}\]Log in or create a free account to answer this question.
