| The reaction \[A\to B\] follows first order kinetics. The time taken for 0.8 mole of A to produce 0.6 mole of B is 1 hour. |
| What is the time taken for conversion of 0.9 mole of A to produce 0.675 mole of B? [AIPMT 2003] |
A) 0.25 h
B) 2 h
C) 1 h
D) 0.5 h
Correct Answer: C
Solution :
| [c] Rate constant of first order reaction |
| \[(k)=\frac{2.303}{t}\,{{\log }_{10}}\frac{{{(A)}_{0}}}{{{(A)}_{t}}}\] |
| or \[k=\frac{2.303}{1}\times {{\log }_{10}}\frac{0.8}{0.2}\] (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
