Given, first order reaction graph between ln(a−x) on Y-axis and t on X-axis.According to first order reaction, ln(a−x)
Rate constant is given byk=tln[a−x][a] or =ln[a−x][a]kt=ln[a]−ln[a−x]or ln[a−x]=ln[a]−ktCompare Eq. (ii) by straight line equation.Y=ln[a−x],x=t,m( slope )=−kC (intercent) =ln al C (intercept) =ln[a] Therefore, slope = rate constant =−(−10−2)=10−2... (iii) Now initial concentration (time t=0 ) Eq. (ii) becomesln[a−x]=ln[a]From graph at t=0,ln[a−x]=−2.303−2.303=ln[a]Initial concentration, [a]=10−1
