Question

Consider the following 2 - step reaction A + B2 + C → AB + BC (slow) A + BC → AB + C (fast) which of the following statements is/are correct?

1. The rate law is third order.

2. BC is a reaction intermediate.

3. C is a catalyst.

A. none

B. 2 & 3 only

C. 1 & 2 only

D. 1 & 3 only

E. all

Answer 1

A + B2 tc - AB + BC (slow) AB+C (fast) A+BC - order of reaction:- m ~ () It is defined as the sum of powers raised over concentrations of reactants in rate Law of reaction which is obtained experimentally. (2) order of a reaction may be zero, Integer , fractional or even negative. for example Rate =kCAJ 3 = K[A]²6] = KCAJ [B] [CJ These Rates all are Third order. - * we write rate law from the slow step reaction (if more than one steps present)

20 * A + B2+C → AB + BC (slow) A TBC - ABtc (fast) we write rate law from show step: we can write rate law from the reaction like Date= K (J' [Bv] 'Ca]' Here all reactante present imole 1+1+1 So, Ordesa cu orderb) a 3. . Third order reaction. * In above reaction 'c' is the reactant only not a catalyst. * In 1st step 'Bc' is formed again 'Bc is Involved in and reaction. so this is intermediate * Ang! - 0. The rate law is third order. © Be ºf a reaction Intermediate so Tang Ć 1&2 only. I