Question

In a given chemical reaction A+2B gives 3C+4D if velocity with respect to C is 1.5 multiple 10^-5. Calculate velocity with respect to other constituent

Answer 1

Answer:

See below.

Explanation:

For the given reaction,

$A+2B\rightarrow 3C+4D$

The rates of reaction are in the relation,

$-\dfrac{d\left[ A\right] }{dt}=-\dfrac{1}{2}\dfrac{d\left[ B\right] }{dt}=\dfrac{1}{3}\dfrac{d\left[ C\right] }{dt}=\dfrac{1}{4}\dfrac{d\left[ D\right] }{dt}$

We are given,

$\dfrac{1}{3}\dfrac{d\left[ C\right] }{dt}=1.5\times 10^{-5}$

Hence, rate of reaction of A,

$-\dfrac{d\left[ A\right] }{dt}=1.5\times 10^{-5}$

Rate of reaction of B,

$\begin{aligned}\dfrac{d\left[ B\right] }{dt}=-2\times 1.5\times 10^{-5}\\ =-3\times 10^{-5}\end{aligned}$

Rate of reaction of D,

$\begin{aligned}\dfrac{d\left[ D\right] }{dt}=4\times 1.5\times 10^{-5}\\ =6\times 10^{-5}\end{aligned}$