A) 87.4; 0.126; 39.7 mins
B) 80.6; 0.135; 40.8 mins
C) 90.5; 0.144; 2829 mins
D) 802; 0.135; 26.6 mins
Correct Answer: A
Solution :
We know that \[\frac{2.303}{t}\,log\,\,\left( \frac{a}{a-x} \right)\,\,=\,\,0.023\,\,mi{{n}^{-1}}\] Let the initial concentration of the substance = 100 and the substance decomposed in 1.5 hours \[\left( 90 min \right) =x,\,\,then,\,\,a - x =100 - x\] Substituting these values in \[K=\frac{2.303}{t}\log \,\left( \frac{a}{a-x} \right)\] \[\Rightarrow \,\,\,\,0.0231=\frac{2.303}{90}\,\,\log \,\,\frac{100}{100-x}\] \[\log \,\,\frac{2.303}{t}\log \,\left( \frac{a}{a-x} \right)\,\,\,\,\Rightarrow \,\,x=87.4\] So the fraction decomposed in 1.5 hours \[=\,\,\frac{87.4}{100}\,\,=\,\,\,0.874\] Fraction remaining behind after 1.5 hours \[=1-0.874=0.126\] The time required for \[60%\] of decomposition \[K=\frac{2.303}{t}\log \,\,\left( \frac{a}{a-x} \right)\] \[\Rightarrow \,\,\,\,0.023=\frac{2.303}{t}\log \,\left( \frac{100}{40} \right)\,\] \[t=\frac{2.303}{0.0231}\log \,\left( \frac{10}{4} \right)\,\,=\,\,\,39.7\,\,\min utes\] Since the reaction is of the first order, the time required to complete specific fraction is independent of initial concentration (or pressure). Hence \[60%\] of the reaction will decompose in 39.7 mins in this case as well.
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