Question

A+2B------->3C+2D. The rate if disappearance of B is 1*10^-2 mol L^-1s^-1. What will be - (1) rate of the reaction. (2) rate of change in concentration of A and C.

Answer 1

Answer :

(1) The rate of reaction is, $0.5\times 10^{-2}moleL^{-1}s^{-1}$

(2) The rate of change in concentration of A and C are, $0.5\times 10^{-2}moleL^{-1}s^{-1}$ and $0.16\times 10^{-2}moleL^{-1}s^{-1}$ respectively.

Explanation : Given,

Rate of disappearance of B = $1\times 10^{-2}moleL^{-1}s^{-1}$

The given rate of reaction is,

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

The expression for rate of reaction :

$\text{Rate of disappearance of A}=-\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{2}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{3}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{2}\frac{d[D]}{dt}$

(1) Now we have to calculate the rate of reaction.

As we know that,

$\text{Rate of reaction}=-\frac{d[A]}{dt}=-\frac{1}{2}\frac{d[B]}{dt}=+\frac{1}{3}\frac{d[C]}{dt}=+\frac{1}{2}\frac{d[D]}{dt}$

So,

$\text{Rate of reaction}=-\frac{1}{2}\frac{d[B]}{dt}=-\frac{1}{2}\times 1\times 10^{-2}moleL^{-1}s^{-1}=0.5\times 10^{-2}moleL^{-1}s^{-1}$

(2) Now we have to calculate the rate of change in concentration of A and C.

$\text{Rate of change of concentration A}=-\frac{d[A]}{dt}=\text{Rate of reaction}=0.5\times 10^{-2}moleL^{-1}s^{-1}$

$\text{Rate of change of concentration C}=+\frac{1}{3}\frac{d[C]}{dt}=\frac{1}{3}(\text{Rate of reaction})=\frac{1}{3}(0.5\times 10^{-2}moleL^{-1}s^{-1})=0.16\times 10^{-2}moleL^{-1}s^{-1}$

Answer 2

Heya mate,

Plz refer to the attachment for your answer

Hope it works｡◕‿◕｡