Question

Tetrachloromethane, CCl4 is produced from the substitution reaction between methane and chlorine gas. If the rate of formation of CCl4 is 0.05 mol dm^-3 min^-1, what is the rate of disappearance of chlorine gas?​

Answer 1

The rate of disappearance of chlorine gas : 0.2 mol/dm³

Further explanation

The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.

For reaction :

$\tt aA+bB\rightarrow cC+dD$

The rate reaction :

$\tt -\dfrac{1}{a}\dfrac{d[-A]}{dt}= -\dfrac{1}{b}\dfrac{d[-B]}{dt}=\dfrac{1}{c}\dfrac{d[C]}{dt}=\dfrac{1}{d}\dfrac{d[D]}{dt}$

Reaction for formation CCl₄ :

CH₄+4Cl₂⇒CCl₄+4HCl

From equation, rate of reaction = rate of formation CCl₄ = 0.05 mol/dm³

Rate of formation of  CCl₄  = reaction rate x coefficient of  CCCl₄

0.05 mol/dm³ = reaction rate x 1⇒reaction rate = 0.05 mol/dm³

The rate of disappearance of chlorine gas (Cl₂) :

Rate of disappearance of  Cl₂  = reaction rate x coefficient of  Cl₂

Rate of disappearance of  Cl₂ = 0.05 x 4 = 0.2 mol/dm³