A) \[5\]
B) \[6\]
C) \[7\]
D) \[8\]
Correct Answer: A
Solution :
[a] For second order reaction: \[{{[R]}_{initial}}=0.08\,M;\] \[{{[R]}_{final}}=0.01M\] \[x=0.08-0.01=0.07M\]\[\therefore \,\,\,\,(a-x)=0.08-0.07=0.01M\] \[{{k}_{2}}=\frac{1}{t}\frac{x}{a(a-x)}\] \[=\frac{1}{70\min }\times \frac{0.07M}{0.08M\times 0.01M}\] ?.(i) Now, time required to become concentration \[=0.04M.\] i.e., \[x=0.04\text{ }M\] \[{{k}_{2}}=\frac{1}{t}\times \frac{0.04M}{0.08M\times (0.08-0.04)M}\] From Eqs. (i) and (ii) \[\frac{0.07}{70\times 0.08\times 0.01}=\frac{0.04}{t\times 0.08\times 0.04}\] ?..(ii) \[t=10\min =2x\min \] \[\therefore \,\,\,\,x=5\min \]
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