Question

. In the nuclear industry, chlorine trifluoride is used to prepare uranium hexafluoride, a volatile compound of uranium used in the separation of uranium isotopes. Chlorine trifluoride is prepared by the reaction Cl2 (g) 3F2 (g) ⟶ 2ClF3 (g). Write the equation that relates the rate expressions for this reaction in terms of the disappearance of Cl2 and F2 and the formation of ClF3.

Answer 1

Answer : The the rate expression will be:

`Rate=-(d[Cl_2])/(dt)=-(1)/(3)(d[F_2])/(dt)=+(1)/(2)(d[ClF_3])/(dt)`

Explanation :

The general rate of reaction is,

`aA+bB\rightarrow cC+dD`

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

`\text{Rate of disappearance of A}=-(1)/(a)(d[A])/(dt)`

`\text{Rate of disappearance of B}=-(1)/(b)(d[B])/(dt)`

`\text{Rate of formation of C}=+(1)/(c)(d[C])/(dt)`

`\text{Rate of formation of D}=+(1)/(d)(d[D])/(dt)`

`Rate=-(1)/(a)(d[A])/(dt)=-(1)/(b)(d[B])/(dt)=+(1)/(c)(d[C])/(dt)=+(1)/(d)(d[D])/(dt)`

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

Now we have to determine the rate expressions for this given reaction.

The given balanced equations is:

`Cl_2(g)+3F_2(g)\rightarrow 2ClF_3(g)`

`\text{Rate of disappearance of }Cl_2=-(d[Cl_2])/(dt)`

`\text{Rate of disappearance of }F_2=-(1)/(3)(d[F_2])/(dt)`

`\text{Rate of formation of }ClF_3=+(1)/(2)(d[ClF_3])/(dt)`

Thus, the rate expression will be:

`Rate=-(d[Cl_2])/(dt)=-(1)/(3)(d[F_2])/(dt)=+(1)/(2)(d[ClF_3])/(dt)`