textA(g) xrightarrow textB(g) is a first order reaction. The initial concentration of A is text0 text.2 textmol textL-1. After 10 min, the concentration of B is found to be text0 text.18 textmol textL-1. The rate constant (in min-1) for the reaction is

Question

textA(g) xrightarrow textB(g) is a first order reaction. The initial concentration of A is text0 text.2 textmol textL-1. After 10 min, the concentration of B is found to be text0 text.18 textmol textL-1. The rate constant (in min-1) for the reaction is (A) 0.2303 (B) 2.303 (C) 0.693 (D) 0.01

Tardigrade Admin · Accepted Answer

Given, A(g) xrightarrowB(g) (first order reaction) Initial concentration of A=0.2 text mol text L-1 Concentration of textB=0.18 text mol textL-1 Concentration of A after 10 min =0.2-0.18 =0.02 text mol textL-1 time (r)= 10 min Rate constant, k=(2.303/t) log (a/a-x) =(2.303/10) log (0.2/0.02) =0.2303 min -1

Q. A(g)​B(g) is a first order reaction. The initial concentration of A is 0.2molL−1. After 10 min, the concentration of B is found to be 0.18molL−1. The rate constant (inmin−1) for the reaction is

0.2303

2.303

0.693

0.01

Given, A(g)​B(g) (first order reaction) Initial concentration of A=0.2molL−1 Concentration of B=0.18 molL−1 Concentration of A after 10 min =0.2−0.18 =0.02 mol L−1 time (r)= 10 min Rate constant, k=t2.303​loga−xa​ =102.303​log0.020.2​ =0.2303min−1

Q. $A(g) B(g)$ is a first order reaction. The initial concentration of A is $0 .2 mol L^{- 1} .$ After 10 min, the concentration of B is found to be $0 .18 mol L^{- 1} .$ The rate constant $(in mi n^{- 1})$ for the reaction is