Chemistry, 13.11.2019 05:31 ninilizovtskt

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?

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Answer from: SaliJano8950

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 = $-r_A=4.0\times 10^{-2}M/s$

Rate of appearing component B = $r_B=2.0\times 10^{-2}M/s$

Rate of appearing component C = $r_C=6.0\times 10^{-2}M/s$

The general reaction is:

$aA\rightarrow bB+cC$

As per question the stoichiometry relation will be:

$\frac{-r_A}{a}=\frac{r_B}{b}=\frac{r_C}{c}$

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

$\frac{4.0\times 10^{-2}}{a}=\frac{2.0\times 10^{-2}}{b}=\frac{6.0\times 10^{-2}}{c}$

$\frac{4}{a}=\frac{2}{b}=\frac{6}{c}$

$\frac{2}{a}=\frac{1}{b}=\frac{3}{c}$

Thus, the stoichiometry relation is:

a : b : c = 2 : 1 : 3

Thus, the required equation will be:

$2A\rightarrow B+3C$

Answer from: Quest

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Answer from: Quest

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At a certain time in a reaction, substance a is disappearing at a rate of 4.0×10−2 m/s, substance b...