Question

For the reaction, $3A+2B\to C+D$, the differential rate law can be written as:

• A. $3\frac{d\left[A\right]}{dt}=\frac{d\left[C\right]}{dt}=k\left[A{]}^n\right[B{]}^m$
• B. $-\frac{d\left[A\right]}{dt}=\frac{d\left[C\right]}{dt}=k\left[A{]}^n\right[B{]}^m$
• C. $+\frac{1}{3}\frac{d\left[A\right]}{dt}=\frac{d\left[C\right]}{dt}=k\left[A{]}^n\right[B{]}^m$
• D. $-\frac{1}{3}\frac{d\left[A\right]}{dt}=\frac{d\left[C\right]}{dt}=k\left[A{]}^n\right[B{]}^m$

Answer 1

The correct option is D $-\frac{1}{3}\frac{d\left[A\right]}{dt}=\frac{d\left[C\right]}{dt}=k\left[A{]}^n\right[B{]}^m$

For the reaction, $3A+2B\to C+D$, the differential rate law can be written as $-\frac{1}{3}\frac{d\left[A\right]}{dt}=\frac{d\left[C\right]}{dt}=k\left[A{]}^n\right[B{]}^m$ as:

Rate, $r=-\frac{1}{3}\frac{d\left[A\right]}{dt}=-\frac{1}{2}\frac{d\left[B\right]}{dt}=\frac{d\left[C\right]}{dt}=\frac{d\left[D\right]}{dt}=k\left[A{]}^n\right[B{]}^m$.