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For a chemical reaction $4A+3B\rightarrow6C+9D$,
rate of formation of C is $6\times10^{-2}$ mol L$^{-1}$ s$^{-1}$ and rate of disappearance of A is $4\times10^{-2}$ mol L$^{-1}$ s$^{-1}$. The rate of reaction and amount of B consumed in interval of 10 seconds, respectively will be:
Detailed Solution
$r=-\frac{1}{4}\frac{d[A]}{dt}=\frac{1}{6}\frac{d[C]}{dt}=\frac{1}{6}\times6\times10^{-2}=1\times10^{-2}$ mol L$^{-1}$ s$^{-1}$. $-\frac{d[B]}{dt}=3r=3\times10^{-2}$ mol L$^{-1}$ s$^{-1}$, so B consumed in 10 s $=3\times10^{-2}\times10=30\times10^{-2}$ mol L$^{-1}$.