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
Correct Answer: A
Solution :
\[2A+B\xrightarrow{{}}{{A}_{2}}B\] Rate of reaction \[=-\frac{1}{2}\frac{\Delta [A]}{\Delta t}=-\frac{\Delta [B]}{\Delta t}=\frac{\Delta [{{A}_{2}}B]}{\Delta t}\] \[\because \] \[-\frac{\Delta [B]}{\Delta t}=-\frac{1}{2}\frac{\Delta [A]}{\Delta t}\] \[\therefore \]The reactant B will disappear at half the rate as A will decrease.
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