A) \[0.0693\,mol\,{{L}^{-1}}\,{{\min }^{-1}}\]
B) \[0.0693\times 2.5\,mol\,{{L}^{-1}}\,{{\min }^{-1}}\]
C) \[0.0693\times 5\,mol\,{{L}^{-2}}\,{{\min }^{-1}}\]
D) \[0.0693\times 10\,mol\,{{L}^{-1}}\,{{\min }^{-1}}\]
Correct Answer: B
Solution :
[b] Initial concentration \[=10\,\,mol\,\,{{L}^{-1}}\] \[\therefore \] Cone. after 20 min (two half-lives) \[=2.5\,mol\,\,{{L}^{-1}}\] Now, \[k=\frac{0.693}{{{t}_{1}}/2}=\frac{0.693}{10\min }\] or \[0.0693\,{{\min }^{-1}}\] \[\therefore \] rate \[=k\times \] [reactant] \[=0.0693\times 2.5\,mol\,\,{{L}^{-1}}\,\,{{\min }^{-1}}\]
You need to login to perform this action.
You will be redirected in
3 sec
