A) 1 hr
B) 0.5 hr
C) 0.25 hr
D) 2 hr
Correct Answer: A
Solution :
For the reaction \[2A+B\to 3C+D\] 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 mole of 5 is formed) Suppose \[{{t}_{1}}\] hour is required for changing the con- centration of A from 0.9 mole to 0.675 mole of B. Remaining mole of A = 0.9 - 0.675 = 0.225 \[\therefore \] \[k=\frac{2.303}{t}{{\log }_{10}}\frac{0.9}{0.225}\] (ii) From Eqs (i) and (ii) \[k=\frac{2.303}{t}{{\log }_{10}}\frac{0.8}{0.2}=\frac{2.303}{{{t}_{1}}}{{\log }_{10}}\frac{0.9}{0.225}\] \[2.303\,{{\log }_{10}}4=\frac{2.303}{t}{{\log }_{10}}4\]
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