Question

At a certain time in a reaction, substance A is disappearing at a rate of 4.0×10−2 M/s, substance B is appearing at a rate of 2.0×10−2 M/s, and substance C is appearing at a rate of 6.0×10−2 M/s. Which of the following could be the stoichiometry for the reaction being studied?

Answer 1

Answer : The required equation will be:

`2A\rightarrow B+3C`

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

As we are given that:

Rate of disappearing component A =
`-r_A=4.0* 10^(-2)M/s`

Rate of appearing component B =
`r_B=2.0* 10^(-2)M/s`

Rate of appearing component C =
`r_C=6.0* 10^(-2)M/s`

The general reaction is:

`aA\rightarrow bB+cC`

As per question the stoichiometry relation will be:

`(-r_A)/(a)=(r_B)/(b)=(r_C)/(c)`

Now put the values of rate of following components, we get:

`(4.0* 10^(-2))/(a)=(2.0* 10^(-2))/(b)=(6.0* 10^(-2))/(c)`

`(4)/(a)=(2)/(b)=(6)/(c)`

`(2)/(a)=(1)/(b)=(3)/(c)`

Thus, the stoichiometry relation is:

a : b : c = 2 : 1 : 3

Thus, the required equation will be:

`2A\rightarrow B+3C`