Question

The following data were obtained for the decomposition of acetaldehyde CH3CHO → CH4 + CO Initial Rate (mol L min-1) [CH3CHO] (mol L-4) 0.20 1.00 0.56 2.00 Which one of the following statements is correct? O A. The reaction is second order. O O B. The rate constant is 0.693 mol L-4 min-?. O C. The rate constant is 0.20 mol-1/2 L 1/2 min-?. O O O D. The reaction is first order. O E. The rate constant is 0.20 mol L-4 min-1

Answer 1

Please see solution
Date : Page No. : - CH₂ CHO CH4+co let assume Rate = k[ CH₂ CH 0] where k is constant & n is order. linitial Rate [CH₂ CHOI 0.20 1 0.56 2 on putting value of Rate & [CH₂ CHO] in equation , we get 0.20 = K 11 - 0.56 = K [2] * -up Dividing (iii) with il, we gel 0.56 5272 0.20 2.8 =12 Taking log on both side. log 2.8 - n log² 0.44 - nxoog n = 1.5 order of Reaction. We know Rate = k [ CH₂ CHO] Rate = K [CH₂ C. Ion putting value, we get Ho

Date: Page No. : 0:20 = K [ ] K = 0.20 [ Rate constant] Unit of Rate constant (K) = Rate [CH₂ CHO]n - Conch Time [conch] of Rate constant - (Conch] - Time! Conch is concentration i.e. where mol i So, unit of Rate Constanta molin-Time 1 mol , Time = moi ? Z mint Rate constant (K) = 0.20 mol 2 2 2 min! (c) is correct.