Question

The rate law for the reaction :$2C+D\to A+Eis-\frac{d\left[D\right]}{dt}=k{\left[C\right]}^2\left[D\right]$
If C is present in large excess, the order of the reaction will be:

• A. zero
• B. first
• C. second
• D. third

Answer 1

The correct option is B first

$-\frac{d\left[D\right]}{dt}=k{\left[C\right]}^2\left[D\right]$
rate is second order w.r.t. to C and first order w.r.t. D so
if we take C in excess, only D will affect the rate and it will become a first order reaction.