A) 0.25 h
B) 2 h
C) 1 h
D) 0.5 h
Correct Answer: C
Solution :
Rate constant of first order reaction \[(k)=\frac{2.303}{t}\,{{\log }_{10}}\frac{{{(A)}_{0}}}{{{(A)}_{t}}}\] \[ork=\frac{2.303}{1}\times {{\log }_{10}}\frac{0.8}{0.2}...(i)\] (because 0.6 moles of B is formed) Suppose \[{{t}_{1}}\] hour are required for the change of concentration of A from 0.9 mole of 0.675 mole of B. Remaining mole of A = 0.9 - 0.675 = 0.225 \[\therefore k=\frac{2.303}{{{t}_{1}}}{{\log }_{10}}\frac{0.9}{0.225}....(ii)\] From Eqs. (i) and (ii) \[\frac{2.303}{1}{{\log }_{10}}\frac{0.8}{0.2}=\frac{2.303}{{{t}_{1}}}{{\log }_{10}}\frac{0.9}{0.225}\] \[2.303{{\log }_{10}}4=\frac{2.303}{t}{{\log }_{10}}4\] \[{{t}_{1}}=1\,h\].
You need to login to perform this action.
You will be redirected in
3 sec
