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All reactions are of 1^{\text {st }} order at time \mathrm{l_1\left(l_1>0\right)}
\mathrm{\frac{[P]}{[Q]}=a}, Therefore at time \mathrm{t_2\left(t_2 \geq t_1\right),}
\mathrm{\frac{[Q]}{[c]}=b}. Which of the following is correct?

 

Option: 1

\mathrm{a>b}


Option: 2

\mathrm{a=b}


Option: 3

\mathrm{ a b=0.4}


Option: 4

\mathrm{ a+b=0.4}


Answers (1)

best_answer

 At any time, \mathrm{ \frac{[P]}{[Q]}=\frac{6 k_1}{16 k_1}=\frac{3}{8}=a=0.375}
At any time, \mathrm{\frac{[Q]}{[C]}=\frac{16 k_1}{15 k_1}=\frac{16}{15}=b=1.067}

\mathrm{a b=\frac{3}{8} \times \frac{16}{15}=\frac{2}{5}=0.4}

Posted by

Ritika Jonwal

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