Question

The rate of formation of C in the reaction 2 A + B →2 C + 3 D is 2.7 mol dm−3 s −1 . State the reaction rate, and the rates of formation or consumption of A, B, and D.

Answer 1

Answer:

Rate of reaction =$1.35mol.dm^{-3}.s^{-1}$

Rate of consumption of A = $2.7mol.dm^{-3}.s^{-1}$

Rate of consumption of B = $1.35mol.dm^{-3}.s^{-1}$

Rate of formation of D = $4.15mol.dm^{-3}.s^{-1}$

Explanation:

According to laws of mass action for the given reaction,

$Rate= -\frac{1}{2}\frac{\Delta [A]}{\Delta t}=-\frac{\Delta [B]}{\Delta t}=\frac{1}{2}\frac{\Delta [C]}{\Delta t}=\frac{1}{3}\frac{\Delta [D]}{\Delta t}$

where, $-\frac{\Delta [A]}{\Delta t}$ is rate of consumption of A, $-\frac{\Delta [B]}{\Delta t}$ is rate of consumption of B, $\frac{\Delta [C]}{\Delta t}$ is rate of formation of C and $\frac{\Delta [D]}{\Delta t}$ is rate of formation of D

Here $\frac{\Delta [C]}{\Delta t}=2.7mol.dm^{-3}.s^{-1}$

So, Rate of reaction = $(\frac{1}{2}\times 2.7mol.dm^{-3}.s^{-1})=1.35mol.dm^{-3}.s^{-1}$

Rate of formation of D = $(\frac{3}{2}\times \frac{\Delta [C]}{\Delta t})=(\frac{3}{2}\times 2.7mol.dm^{-3}.s^{-1})=4.15mol.dm^{-3}.s^{-1}$

Rate of consumption of A = $(\frac{2}{2}\times \frac{\Delta [C]}{\Delta t})=(\frac{2}{2}\times 2.7mol.dm^{-3}.s^{-1})=2.7mol.dm^{-3}.s^{-1}$

Rate of consumption of B = $(\frac{1}{2}\times \frac{\Delta [C]}{\Delta t})=(\frac{1}{2}\times 2.7mol.dm^{-3}.s^{-1})=1.35mol.dm^{-3}.s^{-1}$