Question

In a reaction $2A+B\to A_{2}B,$ the reactant $B$ will disappear at

• A. half the rate as $A$ will decrease
• B. the same rate as $A$ will decrease
• C. twice the rate as $A$ will decrease
• D. half the rate as $A_{2}B$ will form

Answer 1

Answer: (A)

$2A+B\to A_{2}B$
Rate of reaction $=-\frac{1}{2}\frac{\Delta [A]}{\Delta t}$
$=-\frac{\Delta [B]}{\Delta t}=\frac{\Delta \left[A_{2}B\right]}{\Delta t}$
$\because -\frac{\Delta [B]}{\Delta t}=-\frac{1}{2}\frac{\Delta [A]}{\Delta t}$
$\therefore$ The reactant $B$ will disappear at half the rate as $A$ will decrease.