Question

For a reversible reaction, net rate is
$\left(\frac{dx}{dt}\right)=k_1[A{]}^2[B{]}^{-1}-k_2[C]$
Hence given reaction is

• A. $2A+\frac{1}{B}\underset{k_2}{\stackrel{k_1}{\rightleftharpoons }}C$
• B. $2A\underset{k_2}{\stackrel{k_1}{\rightleftharpoons }}C+B-1$
• C. $2A-B\underset{k_2}{\stackrel{k_1}{\rightleftharpoons }}C$
• D. None of these

Answer 1

Answer: (C)

$2A+-1B\underset{k_2}{\stackrel{k_1}{\rightleftharpoons }}C$

For a forward reaction, $\left(\frac{dx}{dt}\right)=k_1[A{]}^2[B{]}^{-1}$ and for a backward

reaction, $\left(\frac{dx}{dt}\right)=k_2[C]$ and thus net rate is

${\left(\frac{dx}{dt}\right)}_{net}={\left(\frac{dx}{dt}\right)}_t-{\left(\frac{dx}{dt}\right)}_b=k_1[A{]}^2[B{]}^{-1}-k_2[C]$