Question

For a reversible reaction, net rate is
$\left(\frac{d x}{d t}\right)=k_{1}[A]^{2}[B]^{-1}-k_{2}[C]$
Hence given reaction is

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

Answer 1

Answer: (C)

$2 A +-1 \,B\, \underset{k_{2}}{\stackrel{k_{1}}{\rightleftharpoons}} C$

For a forward reaction, $\left(\frac{d x}{d t}\right)=k_{1}[A]^{2}[B]^{-1}$ and for a backward

reaction, $\left(\frac{d x}{d t}\right)=k_{2}[C]$ and thus net rate is

$\left(\frac{d x}{d t}\right)_{ net }=\left(\frac{d x}{d t}\right)_{t}-\left(\frac{d x}{d t}\right)_{b}=k_{1}[A]^{2}[B]^{-1}-k_{2}[C]$